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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