Sunday, 10 July 2016

physics class11 vectors problems part1



1. A vector A makes an angle of 20° and B makes an angle of 110° with the X-axis. The magnitudes of these vectors are 3 m and 4 m respectively. Find the resultant.
Answer:   It is clear that angle between vectors A and B, θ = 110° -20° = 90°. Hence magnitude of the resultant = √(A2+B2) = √(32+42) =√(9+16) =√25 = 5 m.
If α be the angle between resultant and A, then tanα=B/A= 4/3, that gives α = 53.13°. So the resultant makes an angle 20°+53.13° =73.13° with the X-axis.

2. Let A and B be the two vectors of magnitude 10 units each. If they are inclined to X-axis at an angle 30° and 60° respectively, find the resultant.
Answer:   Angle between the vectors θ= 60°-30° =30°. Hence magnitude of the resultant = √(A²+B²+2ABcosθ) = √(10²+10²+2.10.10.cos30°) =√(200+200.√3/2) =10√(2+√3) = 19.32 units
Since the vectors have equal magnitudes so the resultant bisects the angle θ between them. Angle between resultant and vector A =30°/2 =15°. So angle between resultant and X-axis =15°+30° =45°.

3. Add vectors A, B and C each having magnitude of 100 units and inclined to the X-axis at an angle 45°, 135° and 315° respectively.
Answer:    For clarity. let us draw the given vectors as below,
Refer Question - 3
It is clear that out of the three vectors B and C are equal in magnitude but opposite in direction, so their sum B + C = 0. Hence sum of the three vectors A+B+C = A which has magnitude of 100 units and inclined at 45° to X-axis.  

4. Let a=4i+3j and b=3i+4j. Find the magnitudes of  (a)  a,(b) b,(c) a+band (d) a-b. 
Answer:  (a)  Magnitude of a = √(4²+3²) =√25 =5,
(b)Magnitude of b = √(3²+4²) =√25 =5, 

a+b =7i+7j and a-b =i-j 
(c) Magnitude of a+b = √(7²+7²) =√(49+49) =7√2
(d) Magnitude of a-b = √(1²+1²) =√2      

5. Refer to figure below. Find (a) the magnitude,(b) x and y components and (c) the angle with the X-axis of the resultant of OA, BC and DE.
Fig for Q-5

Answer:   Let us resolve these vectors along x and y axes first.
Magnitude of x-component of OA = OA. cos 30° =2*√3/2 =√3 m
Magnitude of x-component of BC = BC. cos 120° =1.5 *(-1/2) =-0.75 m
Magnitude of x-component of DE = DE. cos 270° =0 m
(angles of cosine are taken from the positive direction of x-axis)
Magnitude of y-component of OA = OA. sin 30° =2*1/2 =1 m
Magnitude of y-component of BC = BC. sin 120° =1.5 *(√3/2) =0.75√3
Magnitude of y-component of DE = DE. sin 270° =-1 m 
(b)Now x-component of the resultant = Sum of x-components of the vectors =√3-0.75+0 =1.73-0.75 =0.98 m and  y-component of the resultant = Sum of y-components of the vectors =1+0.75√3-1 =1.30 m 
(a) Magnitude of resultant =√(square of x-comp.+square of y-comp) =√(0.98²+1.3²) =√(0.96+1.69) =1.6 m
(c) angle with X-axis of resultant is given by tanθ =Magnitude of y-comp/Magnitude of x-comp = 1.30/0.98 =1.32, so θ=tan-1(1.32).

6. Two vectors have magnitudes 3 units and 4 units respectively. What should be the angle between them if the magnitude of the resultant is (a) 1 unit, (b) 5 units and (c) 7 units.
Answer:   Magnitudes of vectors =a =3 units, b =4 units, let θ be angle between them and c= magnitude of resultant.
We know that c²=a²+b²+2abCosθ, 
So Cosθ=(c²-a²-b²)/2ab, 
(a) If c =1, Cosθ =(1-9-16)/24 =-1, So θ =180°
(b) If c =5, Cosθ =(25-9-16)/24 =0, So θ =90°
(c) If c =7, Cosθ =(49-9-16)/24 =1, So θ =0°
        
7. A spy report about a suspected car reads as follows. "The car moved 2.00 km towards east, made a perpendicular left turn, ran for 500 m, made a perpendicular right turn, ran for 4.00 km and stopped". Find the displacement of the car.
Answer:   Let us take east as X-axis and north as Y-axis and draw the displacements of the car as in figure below, 
Figure for question no.-7
  Taking unit vectors in X and Y directions as i and j, the given displacements can be written as,
OA = 2i, AB =0.5j and BC =4i  
Hence the resultant displacement vector OC = OA+AB+BC=2i,+0.5j+4i  =6i+0.5j  
Magnitude of the displacement = √(6²+0.5²) =√36.25 =  6.02 km
Angle between resultant displacement and east is given by tanθ =(0.5/6) =1/12, 
So θ=tan-1(1/12)

8.  A carrom board  (4 ft x 4 ft square) has the queen at the center. The queen, hit by the striker moves to the front edge, rebounds and goes in the hole behind the striking line. Find the magnitude of displacement of the queen (a) from the center to the front edge, (b) from the front edge to the hole and (c) from the center to the hole. 
Answer:   Let us draw the sketch,
Fig. Q no.- 8
O is the center of the carrom board. P is the point on the front edge where queen strikes. PA is the rebound path. We must understand that like the reflection of light OP and PA will make equal angle θ with the normal at P. From the figure PQ=OQ tan θ =2tan θ and 
DP=DA tanθ =4 tanθ,
DQ =DP+PQ = 2 tanθ + 4 tanθ =6 tan θ = 2,
tan θ =1/3
Now Magnitude of displacement of the queen from.
 (a) the center to the front edge OP = √(OQ²+PQ²) =√(4+4 tan²θ) =2√(1+1/9) =2√(10/9) =⅔√10 ft
 (b) the front edge to hole =PA =√(PD²+DA²) =√(16 tan²θ+16) =(4√10)/3 ft
 (c) the center to the hole OA =√(2²+2²) =√8 =2√2 ft

9. A mosquito net over a 7 ft x 4 ft bed is three ft high. The net has a hole at one corner of the bed through which a mosquito enters the net. It flies and sits at the diagonally opposite upper corner of the net. (a) Find the magnitude of the displacement of the mosquito. (b) Taking the hole as the origin, the length of the bed as the X-axis, its width as the Y-axis and vertically up as the Z-axis, write the components of the displacement vector.
Answer:   Let us see this problem in the following figure:-
Figure for Question no -9
(a) The mosquito net is a cuboid of dimension 7 ft x 4 ft x 3 ft and the magnitude of displacement vector OA of the mosquito is the length of the diagonal of the cuboid which is = √(7²+4²+3²) =√74 ft.  
(b) From the figure magnitudes of the components of displacement vector are 7 ft, 4 ft and 3 ft along X,Y and Z-axes respectively.
  

10. Suppose a is a vector of magnitude 4.5 units due north. What is the vector (a) 3a, (b) -4a ? 
Answer:   (a) Vector 3a is such a vector which magnitude is three times that of magnitude of vector a =3x4.5 =13.5 units and has the same direction, due north.  
(b) The vector -4a is due south with magnitude 4 times that of a =4x4.5 =18 units. 

11. Two vectors have magnitudes 2 m and 3 m. The angle between them is 60°. Find (a) The scalar product of the two vectors, (b) The magnitude of their vector product. 
Answer:   Let the magnitudes of two vectors be, a=2 m and b=3 m
θ= angle between the vectors  =60°.
(a) The scalar product of the twovectors is given by abCosθ
= 2.3.Cos60° = 3 m².
(b) The magnitude of their vector product is = abSinθ =2.3.Sin60° =2.3.√3/2 =3√3 m².

12. Let A1A2A3A4A5A6A1 be a regular hexagon. Write the x- components of the vectors represented by the six sides taken in order. Use the fact that the resultant of these six vectors is zero, to prove that 
cos0+cosπ/3+cos2π/3+cos3π/3+cos4π/3+cos5π/3=0.
Use the known cosine values to verify the result. 

Answer:   Let 'a' be the length of each side of the regular hexagon and ibe the unit vector along x-axis.
X=components of the vector A1A2 =aCos0° i=a i 
vector A2A3  =aCos60° i =0.5a i  
vector A3A4 =aCos120° i=-0.5a i 
vector A4A5 =aCos180° i=-a i
vector A5A6=aCos240° i=-0.5a i 
vector A6A1=aCos300° i=0.5a i   
 Since resultant of these vectors are zero hence sum of magnitudes of x-components of these vectors will also be zero ie. 
aCos0°+aCos60°+aCos120°+aCos180°+aCos240°+aCos300°=0
writing the angles in radian and dividing both sides by a we get 
Cos0+Cosπ/3+Cos2π/3+Cos3π/3+Cos4π/3+Cos5π/3=0      

13. Let a=2i+3j+4k and b=3i+4j+5k. Find the angle between them.
Answer:   Magnitudes a=√(2²+3²+4²) =√29, b=√(3²+4²+5²) =√50 
Dot product of these vectors a.b =(2i+3j+4k).(3i+4j+5k) =2.3+3.4+4.5 =6+12+20 =38 
But the value of dot product is also given by abCosθ 
So, abCosθ=38 
Cosθ=38/ab=38/(√29.√50) =38/√1450 
θ=Cos-1(38/√1450)  

14. Prove that A.(AxB)=0.
Answer:   A.(AxB) = A.(ABsinθ)k  =(ABsinθ)A.k = 0 
(Where A and B are magnitudes and θ angle between them. k is the unit vector perpendicular to both A and B. Since A and k are perpendicular hence A.k=0)

15. If A=2i+3j+4k and B=4i+3j+2k. Find AxB. 
Answer:   AxB=(3.2-4.3)i+(4.4-2.2)j+(2.3-3.4)k = -6i+12j-6k 


16. If A,B,C are mutually perpendicular, show that Cx(AxB)=0. Is the converse true? 
Answer:   Let i, j and k be the unit vectors along A, B and C. 
Now Cx(AxB)= Ck x (ABsin90°)k =(ABC) k x k =0
(Since k x k =0)   
Let us see the converse, given (AxB)xC=0,
if θ be angle between A and B and k unit vector perpendicular to bothA an B 
 => ABsinθ k x C = 0 
=> ABsinθ.1.Csinß.u =0 (ß is angle between k and C and u the unit vector along perpendicular to both k and C)
=> This condition gives either θ =0 or ß=0. So they are not mutually perpendicular and the converse is not true.

17. A particle moves on a given straight line with a constant speed v. At a certain time it is at a point P on its straight line path. O is a fixed point. Show that OP x v is independent of the position P.   
Answer:      
Fig. Q no.-17

OP x v = OP.v.sin(180°-θ) k    (k unit vector perpendicular to plain containing OP and v, and 180°-θ is the angle between them)  
=>OP x v = OP.v.sinθ k  = v.(OP.sinθ) k =(v.OQ) k 
This expression is independent of θ or OP and constant because OQ is the perpendicular distance between line and point O. So OP x v does not depend on the position P.

18. The force on a charged particle due to electric and magnetic fields is given by F=qE+qvxB. Suppose E is along the X-axis and B along the Y-axis. In what direction and with what minimum speed v should a positively charged particle be sent so that the net force on it is zero?  
Answer:  Given that F=qE+qvxB=0,
=>          qvxB=-qE.  
=>          -vB i = -E i  (If v is along z-axis θ=90° for v and B)  
=>          v=E/B

19. Give an example for which A.B=C.B but A ≠ C .
Answer:  A.B=C.B 
=>           AB cosθ = CB cosß 
=>           A cosθ = C cosß   
=>           cosθ = (C/A) cosß               -   (necessary condition)
Let C/A=0.5 and ß=60° then cosθ = 0.5 x 0.5 =0.25, This gives θ =75.5°
So if A=2C,  angle between B & C, ß=60° and angle between B & A, θ =75.5° then  A.B=C.B ; is an example.
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Saturday, 12 March 2016

Julius caeser

JULIUS CAESAR
-WILLIAM SHAKESPEARE
The Play Julius Caesar is in five acts. Given below is a brief summary of the events
that happen before Act II Scene II:--
Two Roman tribunes, Flavius and Murellus, see the common people parading in the
streets instead of working in their shops. They demand to know why the men are not
working. A cobbler informs them that the people are celebrating Caesar's victory.
Murellus is infuriated and tells them that Caesar has not defeated an enemy, but
rather has killed the sons of Pompey the Great. Pompey previously ruled Rome
along with Caesar until their alliance fell apart, at which point they went to battle
over the right to rule.
Julius Caesar triumphantly returns to Rome on the festival of Lupercalia, celebrated
on February 15. He is followed by Antony and Brutus and many followers.
A soothsayer approaches Caesar and calls out for attention. Caesar allows him to
speak, and the man tells Caesar to, "Beware the ides of March". Caesar ignores this
warning and calls the man a dreamer.
Brutus remarks to Cassius that he is afraid the people will crown Caesar king.
Cassius then tells Brutus that "Brutus" is just as good a name as "Caesar", and that
both names could just as easily rule Rome.
Brutus, afraid that Caesar will become a king, struggles to decide whether to take
action with Cassius. Casca remains onstage with Brutus and Cassius and tells them
that the three shouts they heard were because Caesar turned down the crown three
times. Apparently Antony offered him the crown three times, and Caesar turned it
down three times.
Casca then adds that the people forgave Caesar and worshipped him even more for
turning away the crown. Cassius informs the audience in a soliloquy that he will fake
several handwritten notes and throw them into Brutus' room in an attempt to make
Brutus think the common people want him to take action against Caesar.
Cassius then arrives and Casca tells him that the senators are planning to make
Caesar a king the next morning. Cassius draws his dagger and threatens to die before
ever allowing Caesar to achieve so much power. Casca shakes hands with Cassius
and they agree to work together to prevent Caesar from seizing power.
Cinna, a co-conspirator, arrives and together they then leave to go throw Cassius'
handwritten notes through Brutus' window. Cassius indicates that he is quite sure
Brutus will join them within the next day.
Brutus is in his garden and has made up his mind that Caesar must be killed. His
reasons are that Caesar is abusing his power and that he has ascended far too
quickly.
Lucius, Brutus' servant, brings him a letter he has found in Brutus' private room.
Brutus interprets the letter as if it were from all of Rome, telling him to slay Caesar
and restore the republic.
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Cassius is further of the opinion that Mark Antony should be killed along with
Caesar, but again Brutus is against the plan, calling it too "bloody." They plan to
commit their murder of Caesar at the Senate at eight o'clock that morning (it is only
three in the morning at this point). However, they are worried that Caesar will not
show up because he has become so superstitious over the past few months.
Decius tells them that he knows how to flatter Caesar, and assures them that he will
convince Caesar to go to the Senate. Cassius and his followers then depart, leaving
Brutus alone.
Caesar, still in his nightgown, is terrified by a dream his wife Calpurnia has had in
which she cried out, "Help, ho! They murder Caesar!" He orders a servant to go to
the priests and have them sacrifice an animal in order to read the entrails for
predictions of the future.
Calpurnia arrives and tells him that he dare not leave the house that day. Caesar acts
brave and tells her that he fears nothing, and that he will die when it is necessary for
him to die. The servant returns and tells him that the sacrificed animal showed a very
bad omen, namely the beast did not have a heart.
Caesar insists on misinterpreting the omens, but Calpurnia begs him to blame her for
his absence from the Senate, to which he finally agrees. However, Decius arrives at
that moment in order to fetch Caesar to the Senate House. Caesar tells him to inform
the Senate that he will not come this day. Decius claims that he will be mocked if he
cannot provide a better reason than that. Caesar then tells him about Calpurnia's
dream, which Decius reinterprets in a positive light.
Decius then overwhelms Caesar's resistance by asking him if the Senate should
dissolve until a better time when Calpurnia has more favourable dreams.
Decius also tempts Caesar by saying that the Senate plans to give the crown to him
and they may change their minds if he does not go.
Caesar tells Calpurnia that he was acting foolishly, and agrees to go to the Senate.
Cassius and the other conspirators arrive at that moment to accompany him to the
Senate. Antony also appears and joins the group of men who then escort Caesar out
of his house.
Caesar takes his seat in the Senate and proceeds to allow Metellus Cimber to petition
him. The man throws himself down at Caesar's feet in order to beg for his brother's
release from banishment, but is ordered to get up.
Caesar tells him that fawning will not win him any favours. At this Brutus comes
forward and pleads for the man's brother. Cassius soon joins him.
Caesar tells them his decision is, "constant as the Northern Star" and that he will not
remove the banishment. Casca kneels and says ―Speak hands for me‖ .Casca first,
and then the other conspirators and Brutus all stab Caesar who falls saying, "Et tu,
Brute? - Then falls Caesar.
Antony arrives and laments the death of Caesar. He begs the murderers, specifically
Brutus, to tell him why Caesar had to be killed. Brutus tells him that Caesar was
destroying the republic and had to be removed from power.
Antony pretends to be convinced by this and asks the conspirators to, "Let each man
render me his bloody hand" He then shakes hands with each of them, naming them
as he shakes the hand.
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Antony quickly recants his agreement with the murderers, and tells Cassius that he
almost joined them after shaking their hands; He asks them if he may have
permission to take the body to the marketplace and show it to the crowds. Brutus
gives him permission to do this, but immediately Cassius pulls Brutus aside and
says, "You know not what you do‖
Brutus decides to give his speech first, and to allow Antony to speak afterwards,
provided that Antony only says positive things about the conspirators. Antony agrees
to this.
Left alone with the body of Caesar, Antony says, "O pardon me, thou bleeding piece
of earth / That I am meek and gentle with these butchers" He continues, with his
speech becoming ever more violent, "Domestic fury and fierce civil strife / Shall
cumber all the parts of Italy"
Brutus tells the masses that he loved Caesar more than any of them, but that he killed
Caesar because he loved Rome more.
Brutus then asks them if they want him to kill himself for his actions, to which the
crowd replies, "Live, Brutus, live, live!"
He lastly begs them listen to Mark Antony speak and to let him depart alone. He
leaves Mark Antony alone to give his oration.
Antony's speech begins with the famous lines, "Friends, Romans, countrymen, lend
me
your ears". His speech continually praises Brutus as "an honourable man" who has
killed Caesar for being ambitious.
He then presents all of the images of Caesar in which Caesar has not been ambitious,
such as when Caesar thrice refused the crown on the day of Lupercal, or when
Caesar filled the Roman treasury with ransom money from victories in war. The
plebeians slowly become convinced that Caesar was not ambitious and that he was
wrongly murdered.
Antony then pulls out Caesar's will and tells them he should not read it to them.
They beg him to read it, and he finally agrees, but puts if off by descending into the
masses and standing next to the body of Caesar.
He shows them the stab wounds and names the conspirators who gave Caesar the
wounds. The crowd starts to surge away in anarchy, crying, "Revenge! About! Seek!
Burn! Fire! Kill! Slay!" Antony stops them and continues speaking.
He finally reads them the will, in which Caesar has given every Roman citizen
seventy five drachmas. The plebeians react in a frenzy of anger against the men who
killed Caesar, and carry away the body.
Antony says, "Now let it work. Mischief, thou art afoot. / Take thou what course
thou wilt‖. He has successfully instigated the mob to mutiny.

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Patol babu film star-short notes

PATOL BABU, FILM STAR
-Satyajit Ray
Personal Satisfaction vs. Financial Rewards is the theme in the chapter, Patol Babu,
Film Star.
Summary: Personal satisfaction is more important than financial rewards, as

depicted in Satyajit Ray's short story, Patol Babu, Film Star. The main
character Patol Babu realized that personal satisfaction could not be measured and
weighed by money, and so he acted in the film out of passion. He felttowards the job
more than the money he would make from the job.
In Patol Babu, Film Star, Satyajit Ray has highlighted the idea that personal
satisfaction is more important than financial rewards. According to him, one does a
job because he is interested in it rather than getting reward from it.
In the story Patol Babu was given only a minor role in the film, as a pedestrian who
was only needed to collide into the main actor Chancal Kumar and uttered a
monosyllable sound "oh." Nevertheless, his passion for the job drove him to work
hard to give the best performance by rehearsing himself. Eventually, he had done a
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terrific job and received praise from Chancal Kumar-"You timed it so well that I
nearly passed out‖, and Barren Mullick praised him saying, "Jolly good! Why, you're

quite an actor!" He felt very pleased with his performance and "a total satisfaction
swept over him." He thought that it did not matter whether he received
any payment or not. "What were twenty rupees when measured against the intense
satisfaction of a small job done with perfection and dedication?" Thus, he did not
wait to collect his payment.
Patol Babu realized that personal satisfaction could not be measured and weighed by
money. To him, personal satisfaction is more crucial than material rewards. Ever
since the beginning, Patol Babu did not act in the film because of the money merely.
In contrast, it was because of his passion towards the job that drove him to act in the
film. "I'll be paid, of course, but that's not the main thing." He also knew that nobody
would appreciate his performance as he is only a minor actor in the film. Even
though Baren Mullick praised him, he would soon forget about it. "But all his labour
and imagination he had put into this one shot--were these people able to appreciate
that?" However, Patol Babu thought that his own satisfaction was more salient. He
had proven his ability and talent in acting and these worked as a sort of motivation to
him.
In the nutshell, personal satisfaction was more important than financial rewards and
this idea work as one of the main issues in the story.
Solved questions:
1What did Nishikandto Ghosh tell Patol Babu one morning?Why?
One morning Nishikandto Ghosh , patol babu‘s neighbour told him that his youngest

brother in law was looking for an actor for film scene. The person was to be around
fifty, short, and bald headed. Since patol babu had acted earlier and was in need of
work‘ he recommended him.


2.Describe the past of Patol Babu as an actor.?
Patol babu had a real passion at one time. He was always in demand in Jatras, in
amateur theatricals, in plays put up by the club in his neighbourhood. There was a
time when people bought tickets especially to see him.
3.What did Patol Babu do for a living after having been retrenched?
Patol Babu opened a variety store. But he had to wind it up after five years. Then he
had a job in Bengali firm but had to give it up due to high handedness of the boss.
Then he remained an insurance salesman for ten years. Of late he has been with a
firm dealing with scrap iron.
4. How did Patol Babu disclose his pleasure for the film role before his wife?
Patol babu disclosed his pleasure for the film role before his wife in talking about his
past. He told her that his first role on the stage had been of a dead soldier. It was
appreciated by all. The chairman of the municipality then gave him a silver medal.
God willing he would rise to fame again after this role.
I) Answer the following questions in a sentence or two each:
1. How did Sosanko convince Patol Babu about the dialogue given to him?
2. Describe the scene Patol has to play in the film.
3. Which incidents prove that Patol was a man of imagination?
4. How did Patol react to the lines given to him?
5. Why did he leave the shooting spot without taking the money?
6. How does he rate the people in the filmdom?
7. What did Nishikandto Ghosh tell Patol Babu one morning? Why?
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8. Describe the past of Patol Babu as an actor.
9. What did Patol Babu do for a living after having been retrenched?
10. How did Patol Babu disclose his pleasure for the film role before hiswife?
11. What did Patol Babu think of Chanchal Kumar after his shoot?
12. What did Jyoti tell Patol Babu about his role before it was to be shot?
13. What did Patol Babu do with the paper on which his dialogue was written?
14. What did he hear standing near the paan shop and how he reacted to it?
15. What had Patol Babu‘s mentor Gogon Pakrashi told him about the actor as an

artist?
16. How was Patol Babu emboldened by his mentor Gogon Pakrashi‘s advice ?
17. What did Patol Babu suggest Baren Mullick to add authenticity to his role
when histurn for shoot came? What happened thereafter?
18. What did Patol Babu feel after the shoot?Why did he go without taking the money?

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A Shady plot-short notes

-Elsie Brown
This humorous ghost story revolves around the three characters namely John
Hallock, his wife Lavinia and the Ghost Helena. Lavinia suspects that her husband
maintains a secret relationship with a girl and she says this with surety as she has
seen him talking with her alone. She decides to break her relationship with him.
Luckily she comes to know from the ghost itself that it is helping him write a ghost
story. Now they reconcile with each other.
Jenkins, a magazine publisher asks the writer to give some story on super natural
things. The writer accepts. The writer has no specialization in the ghost story
writing. As he sits for writing about ghostly things issues like house hold matters
crop up in his mind. Then he hears a voice and starts his conversation with a sheghost
as it were. The ghost says that she has come on his request through Ouija
board to help him write a story. The ghost asks him to avoid calling through Ouija
board. His wife calls him then and the ghost disappears and warns him about their
strike.


His wife asks him why he is sitting in the dark. She brings an Ouija board to know
about historical events to help him write a story. He does not like the idea and asks
her to return the board aback and get something else in its replacement. His wife
stares at him. He tries to write a story, but cannot do it. Then next Saturday he writes
his story. When he comes back, his mind is churning and his house is brightly

lighted. The room is filled with middle-aged women belonging to his wife‘s book
club. They are sitting with Ouija boards. He is scared that his wife may notice the
ghost. When his wife asks him to join the Ouija board, he refuses. Somehow his wife
comes to know from the Ouija board and the board operator namely Laura that John
has betrayed his wife by having relationship with a girl called Helen. She decides to
break the relationship, but luckily she comes to know from John that Helen is
helping him to write a ghost story.
Solved:
I)Short Answer Question:
a) Why did John wish he were dead?
Ans. The ladies gathered at his house hadtroubled him. The chaos created due to the

ladies‘ conversation and the Ouja board disturbed him a lot. Moreover his wife
doubted that he had an affair with Helen and threatened him that she would divorce
him.
II) Long Question Answer:
Describe A Shady Plot as a ghost story, describing the supernatural and the
atmosphere around which it revolves.
Ans. Helen, an owl-eyed and unattractive ghost plays a vital role in the story. She
appears in parts, first her arm, then a leg, a sleeve and at last, a complete woman
stood there. It does not create fear, on the other hand, the ghost appears to be a real
human being, and helps the couple to reunite at the end, instead of creating trouble in
their lives, as ghosts are supposed to do. Ouija boards are mentioned which are
intended to communicate with the spirits of the dead. Ouija boards help to get
answers to their questions. This story is different from other ghost stories as it does
not send a chill down the spine. Here the ghost does not threaten, but creates
inspiration and the best plot for writing a ghost story.
Questions

1. Why does the ghost appear before the writer?
2.How does the writer react when ghost appear before him?
3.Why does the ghost want the writer to stop his guests from using Ouija Board?
4.Why do Helen and other ghosts organize the writer‘s inspiration bureau?
5.What happened when Hallock sat down his desk and waited for getting ideas for
his ghost story?
6.How did the narrator react to the appearance of the ghost?
7.How was this ghost different form the ghosts the narrator used to write about in his
stories?
8.What did the ghost tell about herself which didn‘t sound that she was a ghost?
9.What did the ghost tell the narrator of her present assignment and what for did she
request him?
10.What made Lavinia buy Ouija board? How did the narrator react to her doing
so?What for did Lavinia buy the Ouija board for the narrator? How could he get help
form it?
11.―My wife is never so pretty as when she is doing something she knows I
disapprove of ―What did she do which he unapproved of?
12.What happened when Laura Hinkle worked on the Ouija board?
13.How did the Ouija board make things easier to understand after Miss Hinkle
asked it to explain itself more fully?
14.When did Helen, the ghost appear before the narrator?
15.‗It is all your fault. She glared at me‖
.What was the narrator‘s fault as per the ghost? What for did the narrator blame the
ghost?
14.Describe how Lavinia faced Helen, the ghost?
15.Explain the appropriateness of the title ―A Shady Plot‖
16.How does the narrator rate himself as a ghost story writer?
17.Why does the ghost appear before the writer?
18.How does the writer react when ghost appears before him?
19.Why does the ghost want the writer to stop his guests from using Ouija Board?
20.Why do Helen and other ghosts organize the writer‘s inspiration bureau?


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Friday, 11 March 2016

Arithmetic Progressions : Previous Year's Questions

One Mark Questions 
Q.1      The nth term of an AP is {\bf{7}} - {\bf{4n}}. Find its common difference.
[Delhi 2008]
Sol.
We have
an = 7 – 4n \Rightarrow {a_1} = 7 - 4 \times 1 = 3 [at n = 1]
\Rightarrow {a_2} = 7 - 4 \times 2 =- 1 [at n = 2]
\Rightarrow {a_3} = 7 - 4 \times 3 =- 5 [at n = 3]
Therefore, common difference  = {\rm{ }}-1-3{\rm{ }} = {\rm{ }}-4
Q.2       The nth term of an AP is{\bf{6n}}{\rm{ }} + {\rm{ }}{\bf{2}}. Find its common difference.
[Delhi 2008]
Sol.
We have
{a_n} = {\rm{ }}6n{\rm{ }} + {\rm{ }}2
\Rightarrow {a_1} = 6 \times 1 + 2 = 8
\Rightarrow {a_2} = 6 \times 2 + 2 = 14
\Rightarrow {a_3} = 6 \times 3 + 2 = 20
Therefore, common difference = {a_2} - {a_1}  = 14{\rm{ }}-{\rm{ }}8{\rm{ }} = {\rm{ }}6

Q.3       Write the next term of the AP \sqrt 8 ,\sqrt {18} ,\sqrt {32}
[AI 2008]
Sol.
We have
\sqrt 8 = 2\sqrt 2
\sqrt {18} = 3\sqrt 2
\sqrt {32} = 4\sqrt 2
Therefore, AP is 2\sqrt 2 ,\,3\sqrt 2 ,4\sqrt 2 .........
Therefore, common difference
d = {a_2} - {a_1}  = 3\sqrt 2 - 2\sqrt 2   = \sqrt 2
Therefore, next term  = 4\sqrt 2 + \sqrt 2 = 5\sqrt 2
Q.4       The first term of an AP is p and its common difference is q. Find its 10th term.
[Foreign 2008]
Sol.
We have
{a} = p
d = {a_2} - {a_1} = {a_3} - {a_2} = ..... = q
{a_n} = a + \left( {n - 1} \right)d
\Rightarrow {a_{10}} = a + \left( {10 - 1} \right)d
\Rightarrow {a_{10}} = a + 9d
\Rightarrow {a_{10}} = p + 9q

Q.5       Find the next term of AP \sqrt 2 ,\sqrt 8 ,\sqrt {18}
[Delhi 2008C]
Sol.
We have \sqrt 2 ,\sqrt 8 ,\sqrt {18}
\sqrt 8 = 2\sqrt 2
\sqrt {18} = 3\sqrt 2
So AP becomes \sqrt 2 ,2\sqrt 2 ,3\sqrt 2 ...........
d = {a_2} - {a_1} = 2\sqrt 2 - \sqrt 2 = \sqrt 2
Next term = 3\sqrt 2 + \sqrt 2 = 4\sqrt 2
Q.6 Which term of the AP 21, 18, 15 ………….. is Zero ?
[Delhi 2008C]
Sol.
Here  a = 21, d = 18 - 21 = - 3
Let {a_n} = 0
\Rightarrow a + \left( {n - 1} \right)d = 0
\Rightarrow 21 + \left( {n - 1} \right)\left( { - 3} \right) = 0
\Rightarrow 21 - 3n + 3 = 0
\Rightarrow 24 - 3n = 0
\Rightarrow 24 = 3n
\Rightarrow n = {{24} \over 3} = 8
Then 8th term of AP 21, 18, 15,………………….. is zero.

Q.7       Which term of AP 14, 11, 18, …………… is -1 ?
[AI 2008 C]
Sol.
We have 14, 11, 18,……………
Herea = 14,{\rm{ }}d{\rm{ }} = {\rm{ }}11{\rm{ }}-{\rm{ }}14{\rm{ }} = {\rm{ }} - {\rm{ }}3
Let {a_n} = - 1
\Rightarrow a + \left( {n - 1} \right)d = - 1
\Rightarrow 14 + \left( {n - 1} \right)\left( { - 3} \right) = - 1
\Rightarrow 14 - 3n + 3 = - 1
\Rightarrow 14 - 3n + 3 + 1 = 0
\Rightarrow 18 - 3n = 0
\Rightarrow 3n = 18
\Rightarrow n = {{18} \over 3}= 6
So 6th term of AP 14, 11, ……………….. is  -1.

Q.8       For what value of p are 2p + 1,\,13,\,5p - 3 three consecutive terms of an AP ?
[Delhi 2009]
Sol.
If terms are in AP then
{a_2} - {a_1} = {a_3} - {a_2} = ...........
{a_2} = 13
{a_1} = 2p + 1
{a_3} = 5p - 3
\Rightarrow 13 - \left( {2p + 1} \right) = \left( {5p - 3} \right) - 13
\Rightarrow 13 - 2p - 1 = 5p - 3 - 13
\Rightarrow 28 = 7p
\Rightarrow p = 4
Then for p{\rm{ }} = {\rm{ }}4 these terms are in AP.


Q.9       For what value of k  are the numbers x, 2x + k and 3x+6 three consecutive terms of an AP ?
[Foreign 2009]
Sol.
If the numbers are in AP then
{a_2} - {a_1} = {a_3} - {a_2}
\Rightarrow 2x + k - x = 3x + 6 - 2x - k
\Rightarrow k + k = 6
\Rightarrow k = 3
Then fork{\rm{ }} = {\rm{ }}3 the numbers are three consecutive term of an AP.

Q.10     If the sum of first p terms of an AP is a{p^2} + bp . Find its common difference.
[Delhi 2010]
Sol.
Sp = a{p^2} + bp
{S_1} = a \times {\left( 1 \right)^2} + b \times 1 = a + b = {a_1}.......(1)
{S_2} = a \times {2^2} + b \times 2 = 4a + 2b
\Rightarrow {a_1} + {a_2} = 4a + 2b [Sum of two terms]
a + b + {a_2} = 4a + 2b [using (1)]
\Rightarrow {a_2} = 3a + b
Now d = {a_2} - {a_1}  = \left( {3a + b} \right) - \left( {a + b} \right) = 2a
Then common difference = {\rm{ }}2a.

Q.11     If the sum of first m terms of an AP is 2m2 + 3m, then what is its second term ?
[Foreign 2010]
Sol. 
{S_m} = 2{m^2} + 3m  (Given)
{S_1} = 2 \times {1^2} + 3 \times 1 = 5 = {a_1}................(1)
{S_2} = 2 \times {2^2} + 3 \times 2 = 14
\Rightarrow {a_1} + {a_2} = 14 [Sum of two terms]
\Rightarrow 5 + {a_2} = 14
\Rightarrow {a_2} = 9
Then second term is 9.
Two  Marks QuestionsQ.1      The 6th term of an AP is – 10 and 10th term is -26. Determine the 15th term of AP .
[Delhi 2006]
Sol.
Let first term of AP  = a
Common difference = d
Now {a_6} = - 10 \Rightarrow a + 5d = - 10…………..(1)
{a_{10}} =- 26 \Rightarrow a + 9d = - 26 ……………..(2)
On Subtracting (1) from (2), We get
4d = - 16
\Rightarrow d = - 4
Substituting in (i), we get
a + 5\left( { - 4} \right) =- 10
\Rightarrow a = 10
Now {a_{15}} = a + 14d = 10 + 14 \times \left( { - 4} \right) =- 46
So,{15^{th}} term of an AP is – 46

Q.2      Find the sum of all the natural numbers less than 100 which are divisible by 6.
[AI 2006 ]
Sol.
Natural numbers less than 100 and divisible by 6 are
6, 12, 18 …………………. 96
This is an A.P. with  a{\rm{ }} = {\rm{ }}6,{\rm{ }}d{\rm{ }} = {\rm{ }}6{\rm{ }},{\rm{ }}n{\rm{ }} = {\rm{ }}16{\rm{ }},l = {\rm{ }}96
Since {S_n} = {n \over 2}\left[ {a + l} \right]
\Rightarrow S = {{16} \over 2}\left[ {6 + 96} \right] = 816
Hence ,Sum = 816

Q.3      How many terms are there in an AP whose first term and 6th term are 12 and 8 respectively, and sum of all its term is  120 ?
[Foreign 2006 ]
Sol.
First term =  a   = {\rm{ }} - {\rm{ }}12
Let common difference   = {\rm{ }}d
Now\,{a_6} \Rightarrow a + 5d = 8
\Rightarrow - 12 + 5d = 8
\Rightarrow 5d = 8 + 12
d{\rm{ }} = {\rm{ }}4
{S_n} = 120 = {n \over 2}\left[ {2a + \left( {n - 1} \right)d} \right]
\Rightarrow 120 = {n \over 2}\left[ {2 \times \left( { - 12} \right) + \left( {n - 1} \right) \times 4} \right]
\Rightarrow 120 = n\left( { - 12} \right) + 2n\left( {n - 1} \right)
\Rightarrow 120 = - 12n + 2{n^2} - 2n
\Rightarrow 120 = - 14n + 2{n^2}
\Rightarrow 2{n^2} - 14n - 120 = 0
\Rightarrow {n^2} - 7n - 60 = 0
\Rightarrow {n^2} - 12n + 5n - 60 = 0
\Rightarrow n\left( {n - 12} \right) + 5\left( {n - 12} \right) = 0
\Rightarrow \left( {n + 5} \right)\left( {n - 12} \right) = 0
Eithern{\rm{ }} = {\rm{ }}12 orn{\rm{ }} = {\rm{ }} - 5[Rejecting n=-5]
Therefore, no. of terms  = {\rm{ }}12

Q.4       Using AP, find the sum of all 3-digit natural numbers which are multiples of 7.
[Delhi 2006 C]
Sol.
1st 3-digit number which is multiple of 7  = {\rm{ }}105
2nd 3-digit number which is multiple of 7  = {\rm{ }}112
So, numbers are 105, 112, 119, ………………994
These numbers are in AP
Herea{\rm{ }} = {\rm{ }}105,{\rm{ }}d{\rm{ }} = {\rm{ }}112{\rm{ }}-{\rm{ }}105{\rm{ }} = {\rm{ }}07
Let {a_n} = {\rm{ }}994
\Rightarrow a + \left( {n - 1} \right)d = 994
\Rightarrow 105 + \left( {n - 1} \right)7 = 994
\Rightarrow \left( {n - 1} \right)7 = 889
\Rightarrow n - 1 = {{889} \over 7} = 127
\Rightarrow n = 128
Now\,\,{S_n} = {n \over 2}[a + l]
\Rightarrow {S_{128}} = {{128} \over 2}[105 + 994] = 70336

 Q.5      In an AP the sum of first n terms is {{5{n^2}} \over 2} + {{3n} \over 2}.Find its 20th term.
[AI 2006 C]
Sol.
We have {S_n} = {{5{n^2}} \over 2} + {{3n} \over 2}
\Rightarrow {S_1} = {{5 \times 1} \over 2} + {{3 \times 1} \over 2} = 4 = {a_1}
\Rightarrow {S_2} = {{5 \times {2^2}} \over 2} + {{3 \times 2} \over 2} = 13
\Rightarrow {a_1} + {a_2} = 13 [Sum of two terms]
\Rightarrow 4 + {a_2} = 13 [using equation (i)]
\Rightarrow {a_2} = 9
d = {a_2} - {a_1} = 9 - 4 = 5
Now\,\,{a_{20}} = a + 19d
\Rightarrow {a_{20}} = 4 + 19 \times 5 = 99.

Q.6      Find the sum of first 25 terms of an AP whose nth term is1 - 4n.
[Delhi 2007]
Sol. 
{a_n} = 1 - 4n (Given)
{a_1} = 1 - 4 \times 1 = - 3
{a_2} = 1 - 4 \times 2 = - 7
d = {a_2} - {a_1} = 7 - \left( { - 3} \right) =- 4
{a_{25}} = a + 24d = - 3 + 24 \times \left( { - 4} \right) = - 99 = l
Since Sn = {n \over 2}\left( {a + l} \right)
\Rightarrow {S_{25}} = {{25} \over 2}\left( { - 3 - 99} \right)
= 25 \times \left( { - 51} \right) = - 1275
Q.7      Which term of the AP 3, 15, 27, 39 ………………… will be 132 more than its 54th term ?
[Delhi 2007]
Sol.
a{\rm{ }} = {\rm{ }}3,{\rm{ }}d{\rm{ }} = {\rm{ }}15-{\rm{ }}3{\rm{ }} = 12
Since {a_n} = a + \left( {n - 1} \right)d
\Rightarrow {a_{54}} = a + \left( {54 - 1} \right)d
\Rightarrow {a_{54}} = 3 + 53 \times 12 = 639
Let an is 132 more than 54th term
Therefore,{a_n} = 639 + 132
\Rightarrow a + \left( {n - 1} \right)d = 771
\Rightarrow 3 + \left( {n - 1} \right) \times 12 = 771
\Rightarrow \left( {n - 1} \right) = {{768} \over {12}}
\Rightarrow n - 1 = 64
\Rightarrow n = 65
So, 65th term is 132 more than 54th term.

Q.8      The first term, common difference and last term of an AP are 12, 6 and 252 respectively.  Find the sum of all terms of this AP .
[AI 2007]
Sol.
Givena{\rm{ }} = {\rm{ }}12,{\rm{ }}d{\rm{ }} = {\rm{ }}6,{\rm{ }}l{\rm{ }} = {\rm{ }}252
\Rightarrow a + \left( {n - 1} \right)d = 252
\Rightarrow 12 + \left( {n - 1} \right) \times 6 = 252
\Rightarrow \left( {n - 1} \right) \times 6 = 240
\Rightarrow n - 1 = {{240} \over 6}
\Rightarrow n = 40 + 1 = 41
Since  {S_n} = {n \over 2}\left[ {a + l} \right]
\Rightarrow {S_n} = {{41} \over 2} \times \left[ {12 + 252} \right]
= {{41} \over 2} \times 264 = 5412

Q.9       Which term of AP 3, 15, 27, 39 …… will be 120 more than 21st term ?
[AI 2009 ]
Sol.
Given a = 3
d = {\rm{ }}15{\rm{ }}-{\rm{ }}3{\rm{ }} = {\rm{ }}12
Since {a_n} = a + (n - 1)d
\Rightarrow {a_{21}} = a + (21 - 1)d = 12 + 20 \times 12 = 243
Let\,{a_m} = {a_{21}} + 120 = 243 + 120 = 363
\Rightarrow a + \left( {m - 1} \right)d = 363
\Rightarrow 3 + \left( {m - 1} \right) \times 12 = 363
\Rightarrow \left( {m - 1} \right) \times 12 = 360
\Rightarrow m - 1 = {{360} \over {12}}
\Rightarrow m = 31
Therefore ,31st term is 120 more than 21st term.

Q.10       Find the common difference of an AP whose first term is 4, last term is 49 and the sum of all its terms is 265 .
[AI 2010]
Sol.
Given  a = 4,l = 49\,and\,{S_n} = 265
\Rightarrow 265 = {n \over 2}\left[ {4 + 49} \right]
\Rightarrow 530 = 53n
\Rightarrow n = 10
Therefore,{\rm{ }}l{\rm{ }} = {\rm{ }}{a_{10}} = {\rm{ }}d{\rm{ }} + {\rm{ }}9d
\Rightarrow 49{\rm{ }} = 4{\rm{ }} + 9{\rm{ }} \times d
\Rightarrow {{45} \over 9} = d
\Rightarrow d = 5
Therefore, common difference  = {\rm{ }}5

Q.11      In an AP, the first term is -4, the last term is 29 and the sum of all its terms is 150. Find its common difference.
[Foreign 2010]
Sol.
Given a = {\rm{ }}-4,{\rm{ }}l{\rm{ }} = {\rm{ }}29,{\rm{ }}{S_n} = {\rm{ }}150
\Rightarrow 150 = {n \over 2}\left[ {a + l} \right]
\Rightarrow 300 = n\left( { - 4 + 29} \right)
\Rightarrow n = {{300} \over {25}} = 12
Therefore, {a_{12}} = 29 = a + 11d
\Rightarrow - 4 + 11d = 29
\Rightarrow 11d = 33
\Rightarrow d = 3
Therefore, common difference = {\rm{ }}3
Q.12      Is -150 is a term of the AP 17, 12, 7, 2 ………..?
[Delhi 2011]
Sol.
Given 17, 12, 7, 2……………
Here a{\rm{ }} = {\rm{ }}17,{\rm{ }}d{\rm{ }} = {\rm{ }}12{\rm{ }}-{\rm{ }}17{\rm{ }} = {\rm{ }} - 5
{a_n} = - 150
\Rightarrow a + \left( {n - 1} \right)d = - 150
\Rightarrow 17 + \left( {n - 1} \right)\left( { - 5} \right) = - 150
\Rightarrow \left( {n - 1} \right)\left( { - 5} \right) = - 167
\Rightarrow n - 1 = {{ - 167} \over { - 5}}
\Rightarrow n = {{167} \over 5} + 1 = {{172} \over 5} = 34{2 \over 3}
\Rightarrow n is not a whole number
Therefore, – 150 is not a term of the AP.
Q.13     Find the numbers of two digit numbers which are divisible by 6.
[AI 2011]
Sol.
Two digit numbers are divisible by 6 are
12, 18 ………………96
Here a{\rm{ }} = {\rm{ }}12,{\rm{ }}d{\rm{ }} = {\rm{ }}18{\rm{ }}-{\rm{ }}12{\rm{ }} = {\rm{ }}d,{\rm{ }}l{\rm{ }} = {\rm{ }}96
Because{a_n} = {\rm{ }}96
\Rightarrow a + \left( {n - 1} \right)d = 96
\Rightarrow 12 + \left( {n - 1} \right)6 = 96
\Rightarrow 2 + \left( {n - 1} \right){{96} \over 6}
\Rightarrow 2 + n - 1 = 16
\Rightarrow n = 16 - 1
\Rightarrow n = 15

Q.14     In an AP, the first term is 12 and common difference is 6. If the last term of the AP is 252, Find its middle term.
[Foreign 2012]
Sol.
Given  a{\rm{ }} = {\rm{ }}12,{\rm{ }}d{\rm{ }} = {\rm{ }}6,{\rm{ }}l{\rm{ }} = {\rm{ }}252, {a_n} = 252
\Rightarrow a + \left( {n - 1} \right)d = 252
\Rightarrow 12 + \left( {n - 1} \right)6 = 252
\Rightarrow \left( {n - 1} \right)6 = 240
\Rightarrow n - 1 = {{240} \over 6} = 40
\Rightarrow n = 40 + 1 = 41
Now middle term  = {{n + 1} \over 2} = {{41 + 1} \over 2} = 21
Since {a_n} = a + (n - 1)d
\Rightarrow {a_{21}} = a + 20d = 12 + 20 \times 6 = 132
Therefore,  is 132 which is middle term of AP.

Q.15     Find the number of three –digit natural numbers. Which is divisible by 11 .
[Foreign 2013]
Sol.
Three digit natural no. divisible by 11 are 110, 121, 132, ……..990
These terms an AP with a{\rm{ }} = {\rm{ }}110,{\rm{ }}d{\rm{ }} = {\rm{ }}11,{\rm{ }}l{\rm{ }} = {\rm{ }}990
As {a_n} = 990
\Rightarrow a + \left( {n - 1} \right)d = 990
\Rightarrow 110 + \left( {n - 1} \right)11 = 990
\Rightarrow \left( {n - 1} \right)11 = 880
\Rightarrow n - 1 = 80
\Rightarrow n = 80 + 1 = 81
Therefore, there are 81 three digit natural numbers.

Three Marks Questions 
Q.1     The sum of n terms of an AP is {\bf{5}}{{\bf{n}}^{\bf{2}}}-{\rm{ }}{\bf{3n}}. Find the AP, also its 10th term.
[Delhi 2008]
Sol.
Sum of n terms of given AP is {S_n} = {\rm{ }}5{n^2}-{\rm{ }}3n
Therefore, sum of (n –1) terms of given AP is  {S_{n - 1}}
\Rightarrow {S_{n - 1}} = 5\left( {{n^2} - 2n + 1} \right) - 3n + 3
\Rightarrow {S_{n - 1}} = 5{n^2} - 10n + 5 - 3n + 3
\Rightarrow {S_{n - 1}} = 5{n^2} - 13n + 8
Therefore, nth term of AP = {a_n} = {S_n} - {S_{n - 1}} = \left( {5{n^2} - 3n} \right) - \left( {5{n^2} - 13n + 8} \right)
\Rightarrow {a_n} = 10n - 8
Therefore, 1st term of AP  = {\rm{ }}10{\rm{ }} \times {\rm{ }}1{\rm{ }}-{\rm{ }}8{\rm{ }} = {\rm{ }}2
2nd term of AP  = {\rm{ }}10{\rm{ }} \times {\rm{ }}2-{\rm{ }}8{\rm{ }} = {\rm{ }}12
and 3rd term of AP = {\rm{ }}10{\rm{ }} \times {\rm{ }}3{\rm{ }}-{\rm{ }}8{\rm{ }} = {\rm{ }}22
Therefore, required AP  = 2,12,22...................
Therefore, {a_{10}} = 10 \times 10 - 8 = 92

Q.2      Find 10th term from the last of the AP 8, 10, 12 …………. 126
[Delhi 2008]
Sol.
Given 8, 10, 12 ……………… 126
Or 126, 124, 122 ………………….. 12, 10, 8
Therefore, 10th term of the end of the AP 8, 10, 12 ………..  126 = 10th term of AP 126, 124 ………….90, 8
Here a{\rm{ }} = {\rm{ }}126,{\rm{ }}d{\rm{ }} = {\rm{ }}124{\rm{ }}-{\rm{ }}126{\rm{ }} = {\rm{ }}-{\rm{ }}2
Since  {a_n} = a + \left( {n - 1} \right)d
\Rightarrow {a_{10}} = 126 + \left( {10 - 1} \right) \times \left( { - 2} \right)

Q.3      The sum of n terms of an AP is 3{n^2} + 5n. Find the AP. Hence, find its 16th term .
[Delhi 2008]
Sol.
{S_n} = 3{n^2} + 5n
\Rightarrow {S_1} = 3 \times {1^2} + 5 \times 1 = 8
\Rightarrow {a_1} = 8
{S_2} = 3 \times {2^2} + 5 \times 2 = 22
Since,{a_1} + {a_2} = 22  [Sum of two terms]
\Rightarrow 8 + {a_2} = 22  [Using (1)]
\Rightarrow {a_2} = 14
d = {a_2} - {a_1} = 14 - 8 = 6
Therefore, AP is 8, 14, 20, 26…………
Now {a_{16}} = a + 15d = 8 \times 15 \times 6 = 98

Q.4      The sum of the 4th and 8th terms of an AP is 24 and sum of 6th and 10th terms is 44. Find the first three terms of the AP.
[AI 2008]
Sol.
Let 1st term  = {\rm{ }}a
Common difference = d
Therefore, {a_n} = a + \left( {n - 1} \right)d
Given {a_4} + {a_8} = 24
\Rightarrow a + 3d + a + 7d = 24
\Rightarrow 2a + 10d = 24
\Rightarrow a + 5d = 12……………….. (1)
Also, Given
{a_6} + {a_{10}} = 44
\Rightarrow a + 5d + a + 9d = 44
\Rightarrow 2a + 14d = 44
\Rightarrow a + 7d = 22 ……………. (2)
Subtracting (1) from (2),we get
2d = 10
\Rightarrow d = 5
And putting d = 5 in (1) and we get
a{\rm{ }} + {\rm{ }}5{\rm{ }} \times {\rm{ }}5{\rm{ }} = {\rm{ }}12
\Rightarrow a = - 13
Therefore, 1st 3 terms are –13, –8, –3

Q.5      For what value of n are the{{\bf{n}}^{{\bf{th}}}}  term of two APs, 63, 65, 67 ……….. and 3, 10, 17 ………. Equal.
[Foreign 2008]
Sol. 
1st AP is 63, 65, 67
{1^{st}}term{\rm{ }} = {\rm{ }}63,Common difference = {\rm{ }}65{\rm{ }}-{\rm{ }}63{\rm{ }} = 2
{A_n} = 63 + \left( {n - 1} \right)2 = 61 + 2n
2nd AP is 3, 10, 17 …………..
a{\rm{ }} = {\rm{ }}3,{\rm{ }}d{\rm{ }} = {\rm{ }}10{\rm{ }}-{\rm{ }}3{\rm{ }} = {\rm{ }}7
{B_n} = 3 + \left( {n - 1} \right)7 = - 4 + 7n
Given{A_n} = {\rm{ }}{B_n}
\Rightarrow 61 + 2n = - 4 + 7n
\Rightarrow 5n = 65
\Rightarrow n = 13
Therefore, at n=13 both AP’s are equal.


Q.6      If m times  the mth term of an AP is equal to n times its nth term, find the (m + n)th term of the AP.
[Foreign 2008]
Sol. 
Let 1st term  = {\rm{ }}a
Common difference = {\rm{ }}d
Therefore, {a_m} = a + \left( {m - 1} \right)d
And {a_n} = a + \left( {n - 1} \right)d
Given m {a_m} = n\,\,{a_n}
\Rightarrow m\left\{ {a + \left( {m - } \right)d} \right\} = n\left\{ {a + \left( {n - 1} \right)d} \right\}
\Rightarrow {m_a} - {n_a} = n\left( {n - 1} \right)d - m\left( {m - 1} \right)d
\Rightarrow {(m - n)_a} = \left( {{n^2} - n - {m^2} + m} \right)d
\Rightarrow {(m - n)_a} = \left( {n - m} \right)\left( {m + n - 1} \right)d
\Rightarrow a = - \left( {m + n - 1} \right)d
Now {a_{m + n}} = a + \left( {m + n - 1} \right)d
\Rightarrow - \left( {m + n - 1} \right)d + \left( {m + n - 1} \right)d=0
Hence {a_{m + n}} = 0  
Q.7      In an AP the first term is 8,{{\bf{n}}^{{\bf{th}}}} term is 33 and sum to first n terms is 123. Find n and d the common difference.
[Foreign 2008]
Sol.
1st term of the AP is a  = {\rm{ }}8
{a_n} = 33 = l
{S_n} = 123
\Rightarrow {n \over 2}\left[ {a + l} \right] = 123
\Rightarrow n\left( {8 + 33} \right) = 123 \times 2
\Rightarrow n = {{123 \times 2} \over {41}}
\Rightarrow n = 6
Also {a_n} = a + \left( {n - 1} \right)d = 33
\Rightarrow 8 + \left( {6 - 1} \right)d = 33
\Rightarrow 5d = 33 - 8
\Rightarrow d = 5
Therefore, common difference is 5 .

Q.8       The first and last terms of an AP are 4 and 81 respectively. If the common difference is 7, how many terms are there in the AP and what their sum ?
[Delhi 2008 C]
Sol.
Given,a{\rm{ }} = {\rm{ }}4,{\rm{ }}{a_n} = {\rm{ }}81andd{\rm{ }} = {\rm{ }}7
Since {a_n} = a + \left( {n - 1} \right)d
\Rightarrow a + \left( {n - 1} \right)d = 81
\Rightarrow 4 + \left( {n - 1} \right)7 = 81
\Rightarrow 4 + 7n - 7 = 81
\Rightarrow 7n = 84
\Rightarrow n = 12
Therefore, there are 12 terms in given AP
Since {S_n} = {n \over 2}\left( {a + l} \right)
\Rightarrow {S_{12}} = {{12} \over 2}\left( {4 + 81} \right) = 6 \times 85 = 510
Therefore, sum of 12 terms is 510.
Q.9       If the sum of first 7 terms of an AP is 49 and that of first 17 terms is 289. Find the sum of first n terms.
[Delhi 2008 C]
Sol.
Let 1st term = a
Common difference = d
Given {S_7} = 49
\Rightarrow {S_7} = {7 \over 2}\left( {2a + 6d} \right)\left[ {Because{S_7} = {7 \over 2}\left( {2a + \left( {7 - 1} \right)} \right)d} \right]
\Rightarrow {7 \over 2}\left( {2a + 6d} \right) = 49
\Rightarrow a + 3d = 7 ……….. (1)
Also, Given{S_{17}} = 289
\Rightarrow {S_{17}} = {{17} \over 2}\left( {2a + 16d} \right)\left[ {Because{S_{17}} = {{17} \over 2}\left( {2a + \left( {17 - 1} \right)} \right)d} \right]
\Rightarrow {{17} \over 2}\left( {2a + 16d} \right) = 289
\Rightarrow a + 8d = 17 …………….. (2)
Subtracting equation (1) from equation (2)we get.
5d = 10
\Rightarrow d = 2
On Puttingd = 2 in equation (1) ,We get
a + 3 \times 2 = 7 \Rightarrow a = 1
Since {S_n} = {n \over 2}\left[ {2a + \left( {n - 1} \right)d} \right]
\Rightarrow {S_n} = {n \over 2}\left[ {2 \times 1 + \left( {n - 1} \right)2} \right] = {n^2}


Q.10      The first and last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there in the AP and what is their sum ?
[AI 2008 C]
Sol.
Givena{\rm{ }} = {\rm{ }}17,{a_n} = {\rm{ }}350,d{\rm{ }} = {\rm{ }}9
\Rightarrow a + \left( {n - 1} \right)d = 350
\Rightarrow 17 + \left( {n - 1} \right)9 = 350
\Rightarrow 9n + 8 = 350
\Rightarrow n = {{342} \over 9} = 38
Therefore, there are 38 terms in given AP
Since {S_n} = {n \over 2}\left( {a + l} \right)
\Rightarrow {S_{38}} = {{38} \over 2}\left[ {17 + 350} \right] = 6973 [Because {a_n} = l]

Q.11     If the 8th term of an AP is 37 and 15th term is 15 more than the 12th term. Find the AP, hence find the sum of the first 15 terms of the AP.
[AI 2008 C]
Sol.
Given {a_8} = a + 7d = 37  …………. (1)
And {a_{15}} - {a_{12}} = 15
\Rightarrow a + 14d - a - 11d = 15
\Rightarrow 3d = 15
\Rightarrow d = 5
Putting d = 5 in equation (1),We get
a + 7 \times 5 = 37
\Rightarrow a = 2
Therefore, AP is 2, 7, 12, 17
Since {S_n} = {n \over 2}\left[ {2a + (n - 1)d} \right]
\Rightarrow {S_{15}} = {{15} \over 2}\left[ {2(2) + \left( {15 - 1} \right)5} \right]
\Rightarrow {S_{15}} = {{15} \over 2}\left[ {2 \times 2 + 14 \times 5} \right] = 15[2 + 7 \times 5]
\Rightarrow {S_{15}} = 30 + 35 \times 15 = 30 + 525 = 555
Q.12    The sum of first six terms of an AP is 42. The ratio of its 10th term to its 30th term is 1 : 3. Calculate the first and the 13th terms of the AP.
[AI 2009]
Sol.
Let first term of AP = a
Common difference = d
Then {S_6} = {6 \over 2}\left\{ {2a + \left( {6 - 1} \right)d} \right\}
\Rightarrow 42 = 3\left( {2a + 5d} \right)
\Rightarrow 2a + 5d = 14 …………..(1)
Also{{{a_{10}}} \over {{a_{30}}}} = {1 \over 3}
\Rightarrow {{a + 9d} \over {a + 29d}} = {1 \over 3}
\Rightarrow 3a + 27d = a + 29d
\Rightarrow 2a = 2d
\Rightarrow a = d………….. (2)
Substituting the value of d in equation (1),We get
2a + 5a = 14
\Rightarrow a = {{14} \over 7} = 2
On Putting a=2 in(2),We get
d = 2
{a_{13}} = a + 12d = 2 + 12 \times 2 = 26
Therefore, 1st term = 2 and 13th term = 26

Q.13    The sum of 5th and 9th term of an AP is 72 and the sum of 7th and 12th terms is 97. Find the AP .
[Delhi 2009]
Sol.
Let 1st term of the AP  = {\rm{ }}a
Common difference  = {\rm{ }}d
Given {a_5} + {a_9} = 72
\Rightarrow a + 4d + a + 8d = 72
\Rightarrow 2a + 12d = 72
\Rightarrow a = 36 - 6d ………….. (1)
Also, Given {a_7} + {a_{12}} = 97
\Rightarrow a + 6d + a + 11d = 97
\Rightarrow 2a + 17d = 97
\Rightarrow 2\left( {36 - 6d} \right) + 17d = 97  [from (1)]
\Rightarrow 72 - 12d + 17d = 97
\Rightarrow 5d = 97 - 72
\Rightarrow 5d = 25
\Rightarrow d = 5
Puttingd{\rm{ }} = {\rm{ }}5 in  equation (1), We get
a = 36 - 6 \times 5 = 6
Therefore, A.P. is 6, 11, 16, 21 ………………

Q.14     If 9th term of an AP is zero. Prove that its 29th term is double of its 19th term.
[Foreign 2009]
Sol.
Let a = first term
d = common difference
Now {a_9} = {\rm{ }}0
\Rightarrow a + 8d = 0
\Rightarrow a = - 8d ……………. (1)
\Rightarrow {a_{29}} = a + 28d = - 8d + 28d   [Using (1)]
\Rightarrow {a_{29}} = 20d…………… (2)
Also {a_{19}} = a + 18d
= - 8d + 18d = 10d ………………… (3)   [Using(1)]
From (2) and (3) we have
{a_{29}} = 2 \times {a_{19}}
Hence proved.

Q.15     In an AP, the sum of first ten terms is – 150 and sum of its next ten terms is – 550. Find the AP.
[Delhi 2010]
Sol.
Let a = first term
d = common difference
Given {S_{10}} = - 150
And {S_{20}} - {S_{10}} = - 550
{S_{20}} = - 550 - 150 = - 700
Therefore,  - 150 = {{10} \over 2}\left[ {2a - 9d} \right]
\Rightarrow 2a + 9d = - 30 ……………… (1)
And  - 700 = {{20} \over 2}[2a + 9d]
\Rightarrow 2a + 19d = - 70 …………… (2)
Solving (1) and (2) ,We get
{ - 10d = 40}
\Rightarrow d = - 4
Putting d = – 4 in (1),We get
2a - 9\left( { - 4} \right) = - 30
\Rightarrow 2a = - 30 + 36
\Rightarrow a = 3
Therefore, AP is 3, -1, -5…….

Q.16     Find the value of the middle term of the following AP :  - 6, - 2 , 2 , …………… 58
[Delhi 2011]
Sol. 
Here  a = {\rm{ }} - {\rm{ }}6,{\rm{ }}d{\rm{ }} = {\rm{ }} - {\rm{ }}2{\rm{ }} + {\rm{ }}6{\rm{ }} = {\rm{ }}4,{\rm{ }}{a_n} = {\rm{ }}58
As {a_n} = {\rm{ }}58
\Rightarrow a + \left( {n - 1} \right)d = 58
\Rightarrow - 6 + \left( {n - 1} \right)4 = 58
\Rightarrow \left( {n - 1} \right)4 = 58 + 6
\Rightarrow \left( {n - 1} \right) = 16
\Rightarrow n = 17
Middle term  = {{n + 1} \over 2} = {{17 + 1} \over 2} = 9
Therefore, 9th term is the middle term.   

Q.17     Determine the AP whose fourth term is 18 and the difference of the nineth term from the fifteenth term is 30 .
[Delhi 2011]
Sol.
Given {a_4} = 18
\Rightarrow a + 3d = 18 ……………. (1)
Also Given  {a_{15}} - {a_9} = 30
\Rightarrow \left( {15 - 9} \right)d = 30
\Rightarrow 6d = 30
\Rightarrow d{\rm{ }} = {\rm{ }}5
Putting value of d in (i), We get
a + 3d = 18
\Rightarrow a + 3 \times 5 = 18
\Rightarrow a = 3
Therefore, required AP is 3, 8, 13 …………….
Q.18     Find the sum of all multiples of 7 lying between 500 and 900.
[AI 2012]
Sol.
First multiple = 504
Second multiple = 511
Last multiple = 896
These multiples are in AP with
a = 504,d = 511 - 504 = 7,{a_n} = l = 896
As {a_n} = 896
\Rightarrow a + \left( {n - 1} \right)d = 896
\Rightarrow 504 + \left( {n - 1} \right)7 = 896
\Rightarrow \left( {n - 1} \right)7 = 392
\Rightarrow n - 1 = 56
\Rightarrow n = 57
Since {S_n} = {n \over 2}[a + l]
\Rightarrow {S_{57}} = {{57} \over 2}\left[ {504 + 896} \right] = {{57} \over 2} \times 1400 = 39900

Q.19     The 19th term of an AP is equal to three times its 6th term. If its 9th term is 19, find the AP.
[AI 2013]
Sol.
Given {a_{19}} = {\rm{ }}3{\rm{ }} \times {\rm{ }}{{\rm{a}}_{\rm{6}}}
Let a = first term
d = common difference
As {a_{19}} = 3 \times {a_6}
\Rightarrow a + 18d = 3 \times \left( {a + 5d} \right)
\Rightarrow a = {3 \over 2}d……………. (1)
Also, Given {a_9} = 19
\Rightarrow a + 8d = 19 ……………. (2)
\Rightarrow {3 \over 2}d + 8d = 19 [using (1)]
\Rightarrow 19d = 38
\Rightarrow d = 2
Putting  d{\rm{ }} = {\rm{ }}2 in equation (1), We get
a = {3 \over 2} \times 2 = 3
Therefore, AP is 3, 5, 7, 9 …………..
( Five Marks Questions )
 Q.1 If the sum of first 4 terms of an AP is 40 and that of first 14 terms 280. Find the sum of its first n terms.
[Delhi 2011]
Sol.
Given {S_4} = {\rm{ }}40  and {S_{14}} = {\rm{ }}280
Sum of n terms is {{\rm{S}}_{\rm{n}}}{\rm{ = }}{{\rm{n}} \over {\rm{2}}}\left[ {{\rm{2a + }}\left( {{\rm{n - 1}}} \right){\rm{d}}} \right]
\Rightarrow {{\rm{S}}_{\rm{4}}}{\rm{ = }}{{\rm{4}} \over {\rm{2}}}\left[ {{\rm{2a + }}\left( {{\rm{4 - 1}}} \right){\rm{d}}} \right]{\rm{ = 40}}
\Rightarrow 2\left[ {2a + 3d} \right] = 40
\Rightarrow 2a + 3d = 20 ……….. (1)
And {S_{14}} = {{14} \over 2}\left[ {2a + \left( {14 - 1} \right)d} \right] = 280
\Rightarrow 7\left( {2a + 13d} \right) = 280
\Rightarrow 2a + 13d = 40 ………. (2)
Subtracting (1) and from (2), We get
10d = 20
\Rightarrow d = 2
Putting d= 2 in equation (1) we get
2a + 3 \times 2 = 20
\Rightarrow 2a + 6 = 20
\Rightarrow 2a = 14
\Rightarrow a = 7
Since {S_n} = {n \over 2}\left[ {2a + \left( {n - 1} \right)d} \right] = {n \over 2}\left[ {2 \times 7 + \left( {n - 1} \right)2} \right]
\Rightarrow {S_n} = {n \over 2}\left( {14 + 2n - 2} \right) = {n \over 2}\left( {2n + 12} \right) = n\left( {n + 6} \right)

Q.2      The first and the last terms of an AP are 8 and 350 respectively. If its common difference is 9, how many terms are there and what is their sum ?
[AI 2011]
Sol. 
Here a{\rm{ }} = {\rm{ }}8,{\rm{ }}l{\rm{ }} = {\rm{ }}350,{\rm{ }}d{\rm{ }} = {\rm{ }}9
Because {a_n} = l = 350
\Rightarrow {a_n} = 350
\Rightarrow a\left( {n - 1} \right)d = 350
\Rightarrow 8 + \left( {n - 1} \right)9 = 350
\Rightarrow \left( {n - 1} \right)9 = 350 - 8
\Rightarrow \left( {n - 1} \right)9 = 342
\Rightarrow \left( {n - 1} \right) = {{342} \over 9}
\Rightarrow n - 1 = 38
\Rightarrow n = 39
Since {S_n} = {n \over 2}\left[ {a + l} \right]
\Rightarrow {S_{39}} = {{39} \over 2}\left[ {8 + 350} \right] = 39 \times {{358} \over 2} = 6981             


Q.3      How many multiples of 4 lie between 10 and 250 ? Also find their sum.
[AI 2011]
Sol.
Multiples of 4 between 10 and 250 be 12, 16, 20 ……….. 248
Here a{\rm{ }} = {\rm{ }}12,{\rm{ }}d{\rm{ }} = {\rm{ }}16{\rm{ }}-{\rm{ }}12{\rm{ }} = {\rm{ }}4 and {a_n} = {\rm{ }}248
Since {a_n} = a + (n - 1)d
\Rightarrow a + \left( {n - 1} \right)d = 248
\Rightarrow 12 + \left( {n - 1} \right)4 = 248
\Rightarrow \left( {n - 1} \right)4 = 236
\Rightarrow \left( {n - 1} \right) = {{236} \over 4} = 59
\Rightarrow n = 59 + 1 = 60
S ince {S_n} = {n \over 2}\left[ {a + l} \right]
\Rightarrow {S_{60}} = {{60} \over 2}\left[ {12 + 248} \right] = 30 \times 260 = 7800
Then, sum of all terms is 7800.

Q.4      Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.
[AI 2012]
Sol.
Let first term of the AP be  {a_1} = a = 5
And Common difference = d
Then , A.T.Q.
\left( {{a_1} + {a_2} + {a_3} + {a_4}} \right) = {1 \over 2}\left( {{a_5} + {a_6} + {a_7} + {a_8}} \right)
As {a_1} = a
Since {a_n} = a + (n - 1)d
\Rightarrow {a_2} = a + \left( {2 - 1} \right)d = a + d
\Rightarrow {a_3} = a + \left( {3 - 1} \right)d = a + 2d
\Rightarrow {a_4} = a + \left( {4 - 1} \right)d = a + 3d
\Rightarrow {a_5} = a + \left( {5 - 1} \right)d = a + 4d
\Rightarrow {a_6} = a + \left( {6 - 1} \right)d = a + 5d
\Rightarrow {a_7} = a + \left( {7 - 1} \right)d = a + 6d
\Rightarrow {a_8} = a + \left( {8 - 1} \right)d = a + 7d
Putting all values in given condition,We get
\left[ {a + \left( {a + d} \right) + \left( {a + 2d} \right) + \left( {a + 3d} \right)} \right] = {1 \over 2}\left[ {\left( {a + 4d} \right) + \left( {a + 5d} \right) + \left( {a + 6d} \right) + \left( {a + 7d} \right)} \right]
\Rightarrow \left( {4a + 6d} \right) = {1 \over 2}\left( {4a + 22d} \right)
\Rightarrow 2\left( {4a + 6d} \right) = 4a + 22d
\Rightarrow 4a + 6d = 2a + 11d
\Rightarrow 2a = 5d
\Rightarrow 2 \times 5 = 5d
\Rightarrow d = 2
So, the common difference is 2.

Q.5      If the sum of first 7 terms of an AP is 49 and that of first 17 terms is 289. Find the sum of its first n terms.
[Delhi 2013]
Sol.
Let first term = a
And common difference = d
Given {S_7} = 49
Since {S_n} = {n \over 2}\left[ {2a + \left( {n - 1} \right)d} \right]
\Rightarrow {S_7} = {7 \over 2}\left[ {2a + \left( {7 - 1} \right)d} \right]
\Rightarrow {S_7} = {7 \over 2}\left[ {2a + 6d} \right]
\Rightarrow 49 = {7 \over 2}\left[ {2a + 6d} \right]
\Rightarrow {{49 \times 2} \over 7} = 2a + 6d
\Rightarrow 14 = 2a + 6d
\Rightarrow a + 3d = 7
\Rightarrow a = 7 - 3d ……….. (1)
Also,Given {S_{17}} = 289
\Rightarrow {{17} \over 2}\left( {2a + 16d} \right) = 289
\Rightarrow 2a + 16d = 34
\Rightarrow a + 8d = 17
\Rightarrow \left( {7 - 3d} \right) + 8d = 17  [using (1)]
\Rightarrow 5d = 10
\Rightarrow d = 2
Putting d{\rm{ }} = {\rm{ }}2 in equation (1), We get
a = 7 - 3 \times 2 = 1
Now  {S_n} = {n \over 2}\left[ {2a + \left( {n - 1} \right)d} \right] = {n \over 2}\left[ {2 \times 1 + \left( {n - 1} \right) \times 2} \right]
\Rightarrow {S_n} = {n \over 2}\left[ {2 + 2n - n} \right] = {n \over 2} \times 2n = {n^2}
Therefore, sum is n2.

Q.6      Students of a school thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class in which they are studying e.g., a section of class I will plant 1 tree, section of class II will plant 2 trees and so on till class XII. There are three sections of each class. Find the total number of trees planted by the student of the school. Pollution control is necessary for everybody’s health. Suggest one more role of student in it.
[Foreign 2013]
Sol.
Number of trees planted by class I = {\rm{ }}3{\rm{ }} \times {\rm{ }}1{\rm{ }} = {\rm{ }}3
[according to given condition i.e. each class have 3 sections]
Number of trees planted by class II = {\rm{ }}2{\rm{ }} \times {\rm{ }}3{\rm{ }} = {\rm{ }}6
Number of trees planted by class III  = {\rm{ }}3{\rm{ }} \times {\rm{ }}3{\rm{ }} = 9
So, number of trees planted by class XII = 12{\rm{ }} \times {\rm{ }}3{\rm{ }} = {\rm{ }}36
Total number of trees planted by students
=3+6+9+12+...........................+36  [12 terms]
Since {S_n} = {n \over 2}[2a + (n - 1)d]
\Rightarrow {S_{12}} = {{12} \over 2}\left[ {3 + 36} \right] = 6 \times 39 = 234
Therefore, total number of trees planted by students are 234.
Role of students for everybody’s health.
Multiple choice questions : -
Q.1       In an AP, if d = –2, n = 5 and an = 0, the value of a is 
             (a) 10                (b) 5                       (c) – 8                 (d) 8
[Delhi 2011]
Sol.       
(d)
Given {a_n} = {\rm{ }}0,{\rm{ }}d{\rm{ }} = {\rm{ }}-2{\rm{ }},{\rm{ }}n{\rm{ }} = {\rm{ }}5
Since {a_n} = {\rm{ }}a{\rm{ }} + {\rm{ }}\left( {n{\rm{ }}-{\rm{ }}1} \right)d
\Rightarrow 0 = a + \left( {n - 1} \right)d
\Rightarrow a = 8
\Rightarrow 0 = a + 4 \times \left( { - 2} \right)
\Rightarrow 0 = a - 8
\Rightarrow a = 8

Q.2       If the common difference of an AP is 3, then a20 – a15 is
               (a) 5                       (b) 3                             (c) 15                       (d) 20
[AI 2011]
Sol.
(c)
Given d = 3
Since  {a_n} = {\rm{ }}a{\rm{ }} + {\rm{ }}\left( {n{\rm{ }}-{\rm{ }}1} \right)d
\Rightarrow {a_{20}} = a + \left( {20 - 1} \right)d = a + 19d
And  {a_{15}} = a + \left( {15 - 1} \right)d = a + 14d
Now {a_{20}} - {a_{15}} = a + 19d - \left( {a + 14d} \right) = a + 19d - a + 14d = 5d = 5 \times 3
\Rightarrow {a_{20}}-{\rm{ }}{a_{15}} = {\rm{ }}15

Q.3      If the nth term of an AP is (2n + 1), then the sum of its first three terms is
            (a) 6n + 3              (b) 15                  (c) 12              (d) 21
[AI 2012]

Sol.
(b)
Given {a_n} = 2n + 1
\Rightarrow {a_1} = 2 \times 1 + 1 = 3
\Rightarrow {a_2} = 2 \times 2 + 1 = 5
\Rightarrow {a_3} = 2 \times 3 + 1 = 7
Then {a_1} + {a_2} + {a_3} = 3 + 5 + 7 = 15

Q.4      The next term of the AP \sqrt {18} ,\sqrt {50} ,\sqrt {98} .......... is
             (a) \sqrt {146}       (b) \sqrt {128}     (c) \sqrt {162}     (d) \sqrt {200}
[Foreign 2012]
Sol.
(c)
Here \sqrt {18} = \sqrt {9 \times 2} = 3\sqrt 2
\sqrt {50} = \sqrt {25 \times 2} = 5\sqrt 2
\sqrt {98} = \sqrt {49 \times 2} = 7\sqrt 2
Now a = 3\sqrt 2
d = 5\sqrt 2 - 3\sqrt 2 = 2\sqrt 2
Then next term   = 7\sqrt 2 + 2\sqrt 2 = 9\sqrt 2 =\sqrt {81 \times 2} =\sqrt {162}

Q.5       The common difference of the  A.P. {1 \over p},{{1 - P} \over p},{{1 - 2P} \over p}......... is
             (a) p                 (b) – p                 (c) – 1                  (d) 1
[Delhi 2013]
Sol.
(c)
Given AP is  {1 \over p},{{1 - P} \over p},{{1 - 2P} \over p}...........
Common difference d = {a_2} - {a_1}
Where {a_2} = {{1 - p} \over p} ………….. (1)
And {a_1} = {1 \over p} ………… (2)
On substituting values,We get
d = {{1 - p} \over p} - {1 \over p} = {{1 - p - 1} \over p} = {{ - p} \over p} = - 1

Q.6       The common difference of the  A.P. {1 \over {2r}},{{1 - 3r} \over {2r}},{{1 - 6r} \over {2r}}  is 
             (a) {{ - 1} \over 2}        (b) {1 \over {2r}}      (c) – 3        (d) {{ - 3} \over 2}
[Foreign 2013]
Sol.
(d)
Given AP is {1 \over {2r}},{{1 - 3r} \over {2r}},{{1 - 6r} \over {2r}}.......
{a_1} = {1 \over {2r}}
{a_2} = {{1 - 3r} \over {2r}}
d = {a_2} - {a_1} = {{1 - 3r} \over {2r}} - {1 \over {2r}} = {{1 - 3r - 1} \over {2r}} = {{ - 3r} \over {2r}}
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