1 Marks Question - MATHS STD 11 Science Questions - Vidyadip

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38 questions · self-marked practice — reveal the answer and mark yourself.

Question 11 Mark

If m^th term of an A.P. is n and n^th term is m, then write its p^th term.

Answer

We have,
a_m = a + (m - 1)d = n
⇒ a + (m-1)d = n
and, a_n = a + (n -1)d = m
⇒ a + (n - 1)d = m
Subtracting equation (2) from equation (1), we get
a + (m - 1) d - a - (n - 1)d = n - m
⇒ (m - 1)d - (n - 1)d = n - m
⇒ d(m - 1 - n + 1) = n - m
⇒ d(m - n) = n - m
$\Rightarrow\text{d}=\frac{-(\text{m}-\text{n})}{(\text{m}-\text{n})}$
$\Rightarrow\text{d}=-1$
Putting d = -1 in equation (1), we get
a +(m - 1)(-1) = n
⇒ a - m + 1 = n
⇒ a = n + m - 1
Now,
a_p =a + (p - 1)d
= n + m -1 +(p + 1)(-1)
= n + m - 1 - p + 1
= n + m - p
$\therefore$ a_p = m + n - p
Hence, p^th term is m + n - p.

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Question 21 Mark

Find the sum of the following arithmetic progression:

50, 46, 42, ... to 10 terms.

Answer

50, 46, 42, ... to 10 terms
$\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\text{s}=\frac{10}{2}[2\times50+(10-1)(-4)]$
$=320$

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Question 31 Mark

The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is given by $\frac{\text{l}^2-\text{a}^2}{\text{k}-(\text{l}+\text{a})},$ then $\text{k}=$

S
2S
3S
None of these

Answer

2S

Solution:

Given:

$\text{S}=\frac{\text{n}}{2}(\text{l}+\text{a})$

$\Rightarrow(\text{l}+\text{a})=\frac{2\text{S}}{\text{n}}$

Also, $\text{d}=\frac{\text{l}^2-\text{a}^2}{\text{k}(\text{l}+\text{a})}$

$\Rightarrow\text{d}=\frac{(\text{l}+\text{a})(\text{l}-\text{a})}{\text{k}-(\text{l}+\text{a})}$

$\Rightarrow\text{d}=\frac{[(\text{n}-1)\text{d}]\times\frac{2\text{S}}{\text{n}}}{\text{k}-\frac{2\text{S}}{\text{n}}}$

$\Rightarrow\text{k}-\frac{2\text{S}}{\text{n}}=(\text{n}-1)\frac{2\text{S}}{\text{n}}$

$\Rightarrow\text{k}=\frac{2\text{S}}{\text{n}}(\text{n}-1+1)$

$\Rightarrow\text{k}=2\text{S}$

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Question 41 Mark

If the sum of p terms of an A.P. is q and the sum of q terms is p, then the sum of p + q terms will be:

0
p - q
p + q
-(p + q)

Answer

-(p + q)

Solution:

$\text{S}_\text{p}=\text{q}$

$\Rightarrow\frac{\text{p}}{2}\big\{2\text{a}+(\text{p}-1)\text{d}\big\}=\text{q}$

$2\text{ap}+(\text{p}-1)\text{pd}=2\text{q}\ .....(1)$

$\text{S}_\text{q}=\text{p}$

$\Rightarrow\frac{\text{q}}{2}\big\{2\text{a}+(\text{q}-1)\text{d}\big\}=\text{p}$

$\Rightarrow2\text{aq}+(\text{q}-1)\text{pd}=2\text{p}\ .....(2)$

Multiplying equation (1) by q and equation (2) by p and then solving, we get:

$\text{d}=\frac{-2(\text{p}-\text{q})}{\text{pq}}$

Now, $\text{S}_{\text{p}+\text{q}}=\frac{(\text{p}+\text{q})}{2}[2\text{a}+(\text{p}+\text{q}-1)\text{d}]$

$=\frac{\text{p}}{\text{2}}[2\text{a}+(\text{p}-1)\text{d}+\text{pd}]+\frac{\text{q}}{2}[2\text{a}+(\text{q}-1)\text{d}+\text{pd}]$

$=\text{S}_\text{p}+\frac{\text{pqd}}{2}+\text{S}_\text{q}+\frac{\text{pqd}}{2}$

$=\text{p}+\text{q}+\text{pqd}$

$=\text{p}+\text{q}-\frac{2(\text{p}+\text{q})\text{pq}}{\text{pq}}$

$=-(\text{p}+\text{q})$

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Question 51 Mark

In the arithmetic progression whose common difference is non-zero, the sum of first 3n terms is equal to the sum of next n terms. Then the ratio of the sum of the first 2n terms to the next 2n terms is:

$\frac{1}{5}$
$\frac{2}{3}$
$\frac{3}{4}$
None of these

Answer

$\frac{1}{5}$

Solution:

$\text{S}_{3\text{n}}=\text{S}_{4\text{n}}-\text{S}_{3\text{n}}$

$\Rightarrow2\text{S}_{3\text{n}}=\text{S}_{4\text{n}}$

$\Rightarrow2\times\frac{3\text{n}}{2}\{2\text{a}+(3\text{n}-1)\text{d}\}=\frac{4\text{n}}{2}\{2\text{a}+(4\text{n}-1)\text{d}\}$

$\Rightarrow3\{2\text{a}+(3\text{n}-1)\text{d}\}=2\{2\text{a}+(4\text{n}-1)\text{d}\}$

$\Rightarrow6\text{a}+9\text{nd}-3\text{d}=4\text{a}+8\text{nd}-2\text{d}$

$\Rightarrow2\text{a}+\text{nd}-\text{d}=0$

$\Rightarrow2\text{a}+(\text{n}-1)\text{d}=0\ .....(1)$

Required ratio: $\frac{\text{S}_{2\text{n}}}{\text{S}_{4\text{n}}-\text{S}_{2\text{n}}}$

$\frac{\text{S}_{2\text{n}}}{\text{S}_{4\text{n}}-\text{S}_{2\text{n}}}=\frac{\frac{2n}{2}\{2a+(2\text{n}-1)\text{d}\}}{\frac{\text{n}4}{2}\{2\text{a}+(4\text{n}-1)\text{d}\}-\frac{2\text{n}}{2}\{2\text{a}+(2\text{n}-1)\}\text{d}}$

$=\frac{\text{n}(\text{nd})}{2\text{n}(3\text{nd})-\text{n}(\text{nd})}$

$=\frac{1}{6-1}$

$=\frac{1}{5}$

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Question 61 Mark

If n arithmetic means are inserted between 1 and 31 such that the ratio of the first mean and n^th mean is 3 : 29, then the value of n is:

10
12
13
14

Answer

14

Solution:

The given series is 1, . . . . . . . . . . . , 31

There are n A.M.s between 1 and 31: 1, A₁, A₂, A₃, . . . . ., A_n,

Common difference, $\text{d}=\frac{31-1}{\text{n}+1}=\frac{30}{\text{n}+1}$

Here, we have:

$\frac{\text{A}_1}{\text{A}_\text{n}}=\frac{3}{29}$

$\Rightarrow\frac{1+\text{d}}{1+\text{nd}}=\frac{3}{29}$

$\Rightarrow\frac{1+\frac{30}{\text{n}+1}}{1+\text{n}\times\frac{30}{\text{n}+1}}=\frac{3}{30}$

$\Rightarrow\frac{\text{n}+1+30}{\text{n}+1+30\text{n}}=\frac{3}{29}$

$\Rightarrow\frac{\text{n}+31}{31\text{n}+1}=\frac{3}{29}$

$\Rightarrow29\text{n}+899=93\text{n}+3$

$\Rightarrow64\text{n}=896$

$\Rightarrow\text{n}=14$

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Question 71 Mark

If, S₁ is the sum of an arithmetic progression of 'n' odd number of terms and S₂ the sum of the terms of the series in odd places, then $\frac{\text{S}_1}{\text{S}_2}=$

$\frac{2\text{n}}{\text{n}+1}$
$\frac{\text{n}}{\text{n}+1}$
$\frac{\text{n}+1}{2\text{n}}$
$\frac{\text{n}+1}{\text{n}}$

Answer

$\frac{2\text{n}}{\text{n}+1}$

Solution:

Let n be an odd number.

Given:

S₁ = Sum of odd number of terms

$=\frac{\text{n}}{2}\{2\text{a}+(\text{n}-1)\text{d}\}\ .....(1)$

Since n is odd, the number of odd places $=\frac{\text{n}+1}{2}$

S₂ = Sum of the terms of a series in odd places

$=\frac{\Big(\frac{\text{n}+1}{2}\Big)}{2}\Big\{2\text{a}\Big(\frac{\text{n}+1}{2}-1\Big)2\text{d}\Big\}$

From equations (1) and (2) we have:

$\frac{\text{S}_1}{\text{S}_2}=\frac{\frac{\text{n}}{2}\{2\text{a}+(\text{n}-1)\text{d}\}}{\frac{\text{n}+1}{4}\{2\text{a}+(\text{n}-1)\text{d}\}}$

$=\frac{2\text{n}}{\text{n}+1}$

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Question 81 Mark

If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers, then k =

$\frac{1}{\text{n}}$
$\frac{\text{n}-1}{\text{n}}$
$\frac{\text{n}+1}{2\text{n}}$
$\frac{\text{n}+1}{\text{n}}$

Answer

$\frac{\text{n}+1}{\text{n}}$

Solution:

Given:

Sum of the even natural numbers = k × Sum of the odd natural numbers

$\frac{\text{n}}{2}\{2\text{a}+(\text{n}-1)\text{d}\}=\text{k}\times\frac{\text{n}}{2}\{2\text{a}+(\text{n}-1)\text{d}\}$

$\Rightarrow\{2\times2+(\text{n}-1)2\}=\text{k}\times\{2\times1+(\text{n}-1)2\}$

$\Rightarrow\frac{4+(\text{n}-1)2}{2+(\text{n}-1)2}=\text{k}$

$\Rightarrow\frac{\text{n}+1}{\text{n}}=\text{k}$

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Question 91 Mark

If in an A.P., S_n = n²p and S_m = m²p, where S_r denotes the sum of r terms of the A.P., then S_p is equal to:

$\frac{1}{2}\text{p}^3$
mn p
P³
(m + n)P²

Answer

P³

Solution:

Given:

$\text{S}_\text{n}=\text{n}^2\text{p}$

$\Rightarrow\frac{\text{n}}{2}\{2\text{a}+(\text{n}-1)\text{d}\}=\text{n}^2\text{p}$

$\Rightarrow2\text{a}+(\text{n}-1)\text{d}=2\text{np}$

$\Rightarrow2\text{a}+(2\text{np}-(\text{n}-\text{1}\text{d}))\ .....(1)$

$\text{S}_\text{m}=\text{m}^2\text{p}$

$\Rightarrow\frac{\text{m}}{2}\{2\text{a}+(\text{m}-1)\text{d}\}=\text{m}^2\text{p}$

$\Rightarrow2\text{a}+(\text{m}-1)\text{d}=2\text{mp}$

$\Rightarrow2\text{a}=2\text{mp}-(\text{m}-1)\text{d}\ .....(2)$

From (1) and (2)

$2\text{np}-(\text{n}-1)\text{d}=2\text{mp}-(\text{m}-1)\text{d}$

$\Rightarrow2\text{p}(\text{n}-\text{m})=\text{d}(\text{n}-1-\text{m}+1)$

$\Rightarrow2\text{p}=\text{d}$

Substituting $\text{d}=2\text{p}$ in equation (1), we get:

$\text{a}=\text{p}$

Sum of p terms of the A.P. is given by:

$\frac{\text{p}}{2}\{2\text{a}+(\text{p}-1)\text{d}\}$

$=\frac{\text{p}}{2}\{2\text{p}+(\text{p}-1)2\text{p}\}$

$=\text{p}^3$

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Question 101 Mark

If 7th and 13th terms of an A.P. be 34 and 64 respectively, then its 18th term is:

87
88
89
90

Answer

89

Solution:

a₇ = 34

⇒ a + 6d = 34 .....(1)

Also, a₁₃= 64

⇒ a + 12d = 64 .....(2)

Solving equations (1) and (2). we get:

a = 4 and d = 5

$\therefore$ a₁₈ = a + 17d

= 4 + 17(5)

= 89

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Question 111 Mark

Let S_n denote the sum of n terms of an A.P. whose first term is a. If the common difference d is given by d = S_n − k S_n-₁ + S_n-2, then k =

1
2
3
None of these

Answer

2

Solution:

Let the A.P. be $\text{a},\ \text{a}+\text{d},\ \text{a}+\text{d},\ \text{a}+3\text{d}\ ...$

Given:

$\text{d}=\text{S}_\text{n}-\text{k}\text{S}_{\text{n}-1}+\text{S}_{\text{n}-2}$

For $\text{n}=3,$ we have

$\text{d}=(3\text{a}+3\text{d})-\text{k}(2\text{a}+\text{d})+\text{a}$

$\Rightarrow4\text{a}+2\text{d}=\text{k}(2\text{a}+\text{d})=0$

$\Rightarrow2(2\text{a}+\text{d})=\text{k}(2\text{a}+\text{d})$

$\Rightarrow2=\text{k}$

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Question 121 Mark

The 10th common term between the A.P.s 3, 7, 11, 15, ... and 1, 6, 11, 16, ... is:

191
193
211
None of these

Answer

191

Solution:

As, the common difference of the A.P. 3, 7, 11, 15, ... = 7 - 3 = 4 and the common difference of the A.P. 1, 6, 11, 16, ... = 6 - 1 = 5 And, the common terms of both the A.P.s will be in A.P.

So, the common difference of the A.P. of the common terms, d = LCM (4, 5) = 4 × 5 = its first common term, a = 11

Now, the tenth common term, a₁₀= a + (10 − 1)d

=11 + 9 × 20

= 11 + 180

= 191

Hence, the correct alternative is option (a).

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Question 131 Mark

If the sum of n terms of an A.P. is 2n² + 5 n, then its nth term is

4n − 3
3n − 4
4n + 3
3n + 4

Answer

4n + 3

Solution:

$\text{S}_\text{n}=2\text{n}^2+5\text{n}$

$\text{S}_1=2.1^2+5.1=8$

$\therefore\text{a}_1=7$

$\text{S}_2=2.2^2+5.2=18$

$\therefore\text{a}_1+\text{a}_2=18$

$\Rightarrow\text{a}_2=11$

Common difference, $\text{d}=11 -7 =4$

$\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$

$=7+(\text{n}-1)4$

$=4\text{n}+3$

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Question 141 Mark

If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times the least, then the numbers are:

5, 10, 15, 20
4, 10, 16, 22
3, 7, 11, 15
None of these

Answer

5, 10, 15, 20

Solution:

Let the four numbers in A.P. be as follows:

$\text{a}-2\text{d},\ \text{a}-\text{d},\ \text{a},\text{a}+\text{d}$

Their sum = 50 (Given)

$\Rightarrow(\text{a}-2\text{d})+(\text{a}-\text{d})+\text{a}+(\text{a}+\text{d}=50$

$\Rightarrow2\text{a}-\text{d}=25\ .....(1)$

Also, $(\text{a}+\text{d})=4(\text{a}-2\text{d})$

$\Rightarrow\text{a}+\text{d}=4\text{a}-8\text{d}$

$\Rightarrow3\text{d}=\text{a}\ .....(2)$

From equations (1) and (2),

$\text{d}=5,\ \text{a}=15$

Hence, the numbers are 15 - 10, 15 - 5, 15, 15 + 5, i.e. 5, 10, 15, 20.

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Question 151 Mark

If a₁, a₂, a₃, .... a_n are in A.P. with common difference d, then the sum of the series sin d,
$[\text{cosec}\ \text{a}_1\ \text{cosec}\ \text{a}_2\ \text{cosec}\ \text{a}_1\ \text{cosec}\ \text{a}_3\\+\ .....\ +\text{cosec}\ \text{a}_{\text{n}-1}\text{cosec}\ \text{a}_\text{n}]$ is:

$\sec\text{a}_1-\sec\text{a}_\text{n}$
$\text{cosec}\ \text{a}_1-\text{cosec}\ \text{a}_\text{n}$
$\cot\text{a}_1-\cot\text{a}_\text{n}$
$\tan\text{a}_1-\tan\text{a}_\text{n}$

Answer

$\cot\text{a}_1-\cot\text{a}_\text{n}$

Solution:

We have,

$\sin\text{d}(\text{cosec}\ \text{a}_1\ \text{cosec}\ \text{a}_2+\ \text{coses}\ \text{a}_2\ \text{cosec}\ \text{a}_3\\+\ .....\ +\ \text{cosec}\ \text{a}_{\text{n}-1}\ \text{cosec}\ \text{a}_\text{n})$

$=\frac{\sin\text{d}}{\sin\text{a}_1\sin\text{a}_2}+\frac{\sin\text{d}}{\sin\text{a}_2\sin\text{a}_3}+\ .....\ +\frac{\sin\text{d}}{\sin\text{a}_{\text{n}-1}\sin\text{a}_\text{n}}$

$=\frac{\sin(\text{a}_2-\text{a}_1)}{\sin\text{a}_1\sin\text{a}_2}+\frac{\sin(\text{a}_3-\text{a}_2)}{\sin\text{a}_2\sin\text{a}_3}+\ .....\ +\frac{\sin(\text{a}_\text{n}-\text{a}_{\text{n}-1})}{\sin\text{a}_{\text{n}-1}\sin\text{a}_n}$

$=\frac{\sin\text{a}_2\cos\text{a}_1-\cos\text{a}_2\sin\text{a}_1}{\sin\text{a}_1\sin\text{a}_2}+\frac{\sin\text{a}3\cos\text{a}_2-\cos\text{a}_3\sin\text{a}_2}{\sin\text{a}_1\sin\text{a}_2}\\+\ .....\ +\frac{\sin\text{a}_2\cos\text{a}_1-\cos\text{a}_2\sin\text{a}_1}{\sin\text{a}_1\sin\text{a}_2}$

$=(\cot\text{a}_1-\cot\text{a}_2)+(\cot\text{a}_2-\cot\text{a}_3)\\+\ .....\ +(\cot\text{a}_{\text{n}-1}-\cot\text{a}_\text{n})$

$=\cot\text{a}_1-\cot\text{a}_\text{n}$

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Question 161 Mark

If in an A.P., the pth term is q and (p + q)th term is zero, then the qth term is:

-p
p
p + q
p - q

Answer

p

Solution:

As, $\text{a}_\text{p}=\text{q}$

$\Rightarrow(\text{p}-1)\text{d}=\text{q}\ .....(1)$

Also, $\text{a}_{(\text{p}+\text{q})}=0$

$\Rightarrow\text{a}+(\text{p}+\text{q}-1)\text{d}=0\ .....(2)$

Subtracting (1) from (2), we get

$\text{a}+(\text{p}+\text{q}-1)\text{d}-\text{a}-(\text{p}-1)\text{d}=0-\text{q}$

$\Rightarrow(\text{p}+\text{q}-1)\text{d}-\text{a}-(\text{p}-1)\text{d}=0-\text{q}$

$\Rightarrow\text{pd}=-\text{q}$

$\Rightarrow\text{d}=\frac{-\text{q}}{\text{q}}$

$\Rightarrow\text{d}=-1$

Substituting $\text{d}=-1$ in (1), we get

$\text{a}+(\text{p}-1)\times(-1)=\text{q}$

$\Rightarrow\text{a}-\text{p}+\text{1}=\text{q}$

$\Rightarrow\text{a}=\text{p}+\text{q}-1$

Now,

$\text{a}_\text{q}=\text{a}+(\text{q}-1)\text{d}$

$=\text{p}+\text{q}-1+(\text{q}-1)\times(-1)$

$=\text{p}+\text{q}-1-\text{q}+1$

$=\text{p}$

Hence, the correct alternative is option (b).

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Question 171 Mark

Let S_n denote the sum of first n terms of an A.P. If S_2n = 3 S_n, then S_3n : S_n is equal to:

4
6
8
10

Answer

6

Solution:

As, $\text{S}_{2\text{n}}=3\text{S}_\text{n}$

$\Rightarrow\frac{2\text{a}}{2}[2\text{a}+(2\text{n}-1)\text{d}]=\frac{3\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$

$\Rightarrow2[2\text{a}+(2\text{n}-1)\text{d}]=3[2\text{a}+(\text{n}-1)\text{d}]$

$\Rightarrow4\text{a}+2(2\text{n}-1)\text{d}=6\text{a}+3(\text{n}-1)\text{d}$

$\Rightarrow4\text{a}+\text{nd}4-2\text{d}=6\text{a}+3\text{nd}-3\text{d}$

$\Rightarrow6\text{a}-4\text{a}+3\text{nd}-\text{3d}-4\text{nd}+2\text{d}=0$

$\Rightarrow2\text{a}-\text{nd}-\text{d}=0$

$\Rightarrow2\text{a}-(\text{n}+1)\text{d}=0$

$\Rightarrow2\text{a}=(\text{n}+1)\text{d}\ .....{(1)}$

Now,

$\frac{\text{S}_{3\text{n}}}{\text{S}_\text{n}}=\frac{\frac{3\text{n}}{2}[2\text{a}+(3\text{n}-1)\text{d}]}{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]}$

$=\frac{3[ (\text{n}+1)\text{d}+(3\text{n}-1)\text{d}]}{[(\text{n}+1)+(\text{n}-1)\text{d}]}$ [Using (1)]

$=\frac{3[\text{nd}+\text{d}+3\text{nd}-\text{d}]}{[\text{nd}+\text{d}+\text{nd}-\text{d}]}$

$=\frac{3[4\text{nd}]}{[2\text{nd}]}$

$=3\times2$

$=6$

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Question 181 Mark

If the sum of n terms of an A.P. be 3n² − n and its common difference is 6, then its first term is:

2
3
1
4

Answer

2

Solution:

$\text{S}_\text{n}=3\text{n}^2-\text{n}$

$\Rightarrow\text{S}_1=3(1)^2-1$

$\Rightarrow\text{S}_1=2$

$\therefore\text{a}_1=2$

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Question 191 Mark

In n A.M.'s are introduced between 3 and 17 such that the ratio of the last mean to the first mean is 3 : 1, then the value of n is:

6
8
4
None of these.

Answer

6

Solution:

Let A₁, A₂, A₃, A₄ .... A_n be the n arithmetic means between 3 and 17. Let d be the common difference of the A.P. 3, A₁, A₂, A₃, A₄ .... A_n and 17. Then, we have:

$\text{d}=\frac{17-3}{\text{n}+1}=\frac{14}{\text{n}+1}$

Now, $\text{A}_1=3+\text{d}=3+\frac{14}{\text{n}+1}=\frac{3\text{n}+17}{\text{n}+1}$

And, $\text{A}_\text{n}=3+\text{nd}=3+\text{n}\Big(\frac{14}{\text{n}+1}\Big)=\frac{17\text{n}+3}{\text{n}+1}$

$\therefore\frac{\text{A}_\text{n}}{\text{A}_1}=\frac{\text{3}}{1}$

$\Rightarrow\frac{\Big(\frac{17\text{n}+3}{\text{n}+1}\Big)}{\Big(\frac{3\text{n}+17}{\text{n}+1}\Big)}=\frac{3}{1}$

$\Rightarrow\frac{17\text{n}+3}{3\text{n}+17}=\frac{3}{1}$

$\Rightarrow17\text{n}+3=9\text{n}+51$

$\Rightarrow8\text{n}=48$

$\Rightarrow\text{n}=6$

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Question 201 Mark

If S_n denotes the sum of first n terms of an A.P. < a_n > such that $\frac{\text{S}_\text{m}}{\text{S}_\text{n}}=\frac{\text{m}^2}{\text{n}^2},$ then $\frac{\text{a}_\text{m}}{\text{a}_\text{n}}=$

$\frac{2\text{m}+1}{2\text{n}+1}$
$\frac{1\text{m}-1}{2\text{n}-1}$
$\frac{\text{m}-1}{\text{n}-1}$
$\frac{\text{m}+1}{\text{n}-1}$

Answer

$\frac{1\text{m}-1}{2\text{n}-1}$

Solution:

$\frac{\text{S}_\text{m}}{\text{S}_\text{n}}=\frac{\text{m}^2}{\text{n}^2}$

$\Rightarrow\frac{\frac{\text{m}}{2}\{2\text{a}+(\text{m}-1\text{d})\}}{\frac{\text{n}}{2}\{2\text{a}+(\text{n}-1)\text{d}\}}=\frac{\text{m}^2}{\text{n}^2}$

$\Rightarrow\frac{\{2\text{a}+(\text{m}-1)\text{d}\}}{\{2\text{a}+(\text{n}-1)\text{d}\}}=\frac{\text{m}}{\text{n}}$

$\Rightarrow2\text{an}+\text{ndm}-\text{nd}=2\text{nmd}-\text{md}$

$\Rightarrow2\text{an}-2\text{am}-\text{nd}+\text{md}=0$

$\Rightarrow2\text{a}(\text{n}-\text{m})-\text{d}(\text{n}-\text{m})=0$

$\Rightarrow2\text{a}(\text{n}-\text{m})=\text{d}(\text{n}-\text{m})$

$\Rightarrow\text{d}=2\text{a}\ .....(1)$

Ratio of $\frac{\text{a}_m}{\text{a}_n}=\frac{\text{as}+(\text{m}-1)\text{d}}{\text{a}+(\text{n}-\text{1})\text{d}}$

$\Rightarrow\frac{\text{a}_\text{m}}{\text{a}_n}=\frac{\text{a}+(\text{m}-)2\text{a}}{\text{a}+(\text{n}-)2\text{a}}$ From (1)

$=\frac{\text{a}+2\text{am}-2\text{a}}{\text{a}+2\text{an}-2\text{a}}$

$=\frac{2\text{am}-\text{a}}{2\text{an}-\text{a}}$

$=\frac{2\text{m}-1}{2\text{n}-1}$

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Question 211 Mark

The first and last terms of an A.P. are 1 and 11. If the sum of its terms is 36, then the number of terms will be:

5
6
7
8

Answer

6

Solution:

$\text{a}=1,\ \text{a}_\text{n}=11,\ \text{S}_\text{s}=36$

$\because\text{a}_\text{n}=11$

$\Rightarrow\text{a}+(\text{n}-1)\text{d}=11$

$\Rightarrow1+(\text{n}-1)\text{d}=11$

$\Rightarrow(\text{n}-1)\text{d}=10\ .....(1)$

Also, $\text{S}_\text{n}=36$

$\Rightarrow\frac{\text{n}}{2}\{2\text{a}+(\text{n}-1)\text{d}\}=36$

$\Rightarrow\frac{\text{n}}{2}\{2\text{a}+(\text{n}-1)\text{d}\}=36$

$\Rightarrow\text{n}\{2+10\}=72$ (Uising(1))

$\Rightarrow\text{n}=6$

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Question 221 Mark

If a₁, a₂, a₃, .... a_n are in A.P. with common difference d, then the sum of the series $\sin\text{d}\ [\text{sec}\ \text{a}_1\ \text{sec}\ \text{a}_2\ \text{sec}\ \text{a}_2\ \text{sec}\ \text{a}_3+\ .....\ +\text{sec}\ \text{a}_{\text{n}-1}\text{sec}\ \text{a}_\text{n}]$ is:

$\sec\text{a}_1-\sec\text{a}_\text{n}$
$\text{cosec}\ \text{a}_1-\text{cosec}\ \text{a}_\text{n}$
$\cot\text{a}_1-\cot\text{a}_\text{n}$
$\tan\text{a}_1-\tan\text{a}_\text{n}$

Answer

$\tan\text{a}_1-\tan\text{a}_\text{n}$

Solution:

We have,

$\sin\text{d}(\text{sec}\ \text{a}_1\ \text{sec}\ \text{a}_2+\ \text{ses}\ \text{a}_2\ \text{sec}\ \text{a}_3\\+\ .....\ +\ \text{sec}\ \text{a}_{\text{n}-1}\ \text{sec}\ \text{a}_\text{n})$

$=\frac{\sin\text{d}}{\cos\text{a}_1\cos\text{a}_2}+\frac{\sin\text{d}}{\cos\text{a}_2\cos\text{a}_3}+\ .....\ +\frac{\sin\text{d}}{\cos\text{a}_{\text{n}-1}\cos\text{a}_\text{n}}$

$=\frac{\sin(\text{a}_2-\text{a}_1)}{\cos\text{a}_1\cos\text{a}_2}+\frac{\sin(\text{a}_3-\text{a}_2)}{\cos\text{a}_2\cos\text{a}_3}+\ .....\ +\frac{\sin(\text{a}_\text{n}-\text{a}_{\text{n}-1})}{\cos\text{a}_{\text{n}-1}\cos\text{a}_n}$

$=\frac{\sin\text{a}_2\cos\text{a}_1-\cos\text{a}_2\sin\text{a}_1}{\cos\text{a}_1\cos\text{a}_2}+\frac{\sin{\text{a}_3}\cos\text{a}_2-\cos\text{a}_3\sin\text{a}_2}{\cos\text{a}_1\cos\text{a}_2}\\+\ .....\ +\frac{\sin\text{a}_2\cos\text{a}_1-\cos\text{a}_2\sin\text{a}_1}{\cos\text{a}_1\cos\text{a}_2}$

$=(\tan\text{a}_1-\tan\text{a}_2)+(\tan\text{a}_2-\tan\text{a}_3)\\+\ .....\ +(\tan\text{a}_{\text{n}-1}-\tan\text{a}_\text{n})$

$=\tan\text{a}_1-\tan\text{a}_\text{n}$

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Question 231 Mark

Sum of all two digit numbers which when divided by 4 yield unity as remainder is:

1200
1210
1250
None of these.

Answer

1210

Solution:

The given series is 13, 17, 21 ... 97.

a₁ = 13, a₂ = 17, a_n = 97

d = a₂ - a₁ = 7 - 3 = 4

a_n = 97

⇒ a +(n - 1)d = 97

⇒ 13 + (n - 1)4 = 97

⇒ n = 22

Sum of the above series:

$\text{S}_22=\frac{22}{2}\big\{2\times13+(22-1)4\big\}$

$=11\big\{26+84\big\}$

$=1210$

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Question 241 Mark

If the first, second and last term of an A.P are a, b and 2 a respectively, then its sum is:

$\frac{2\text{n}}{\text{n}+1}$
$\frac{\text{n}}{\text{n}+1}$
$\frac{\text{n}+1}{2\text{n}}$
$\frac{\text{n}+1}{\text{n}}$

Answer

$\frac{\text{n}+1}{2\text{n}}$

Solution:

Let the A.P. be a, a + d, a + 2d ........ a + nd.

Here, let d be the common difference and n be the total number of terms.

Given:

$\text{a}_1=\text{a},$

$\text{a}_2=\text{b}$

$\Rightarrow\text{d}=\text{b}-\text{a}\ .....(1)$

$\text{a}_\text{n}=2\text{a}$

$\Rightarrow\text{a}+(\text{n}-1)\text{d}=2\text{a}$

$\Rightarrow(\text{n}-1)\text{d}=\text{a}$

$\Rightarrow\text{d}=\frac{\text{a}}{\text{n}-1}\ .....{(2)}$

From equations (1) and (2), we have

$\Rightarrow\frac{\text{a}}{\text{n}-1}=\text{b}-\text{a}$

$\Rightarrow\frac{\text{a}}{\text{b}-\text{a}}+1=\text{n}$

$\Rightarrow\frac{\text{a}+\text{b}-\text{a}}{\text{b}-\text{a}}=\text{n}$

$\Rightarrow\frac{\text{b}}{\text{b}-\text{a}}=\text{n}$

Now, sum of n terms of an A.P.

$\text{S}=\frac{\text{n}}{2}\{\text{a}+\text{a}_\text{n}\}$

$=\frac{\text{n}}{2}(3\text{a})$

$=\frac{3\text{ab}}{2(\text{b}-\text{a})}$

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Question 251 Mark

If the sum of n terms of an A.P., is 3n² + 5n then which of its terms is 164?

26th
27th
28th
None of these.

Answer

27th

Solution:

$\text{S}_\text{n}=3\text{n}^2+5\text{n}$

$\text{S}_1=3(1)^2+5(1)=8$

$\therefore\text{a}_1=8$

$\text{S}_2=3(2)^2+5(2)=22$

$\therefore\text{a}_1+\text{a}_2=22$

$\Rightarrow\text{a}_2=14$

Common difference, $\text{d}=14 -8 =6$

Also, $\text{a}_\text{n}=164$

$\Rightarrow\text{a}+(\text{n}-1)\text{d}=164$

$\Rightarrow8+(\text{n}-1)6=164$

$\Rightarrow\text{n}=27$

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Question 261 Mark

If in an A.P. $\text{S}_\text{n}=\text{n}^2\text{q}$ and $\text{S}_\text{m}=\text{m}^2\text{q},$ where Sr denotes the sum of r terms of the A.P., then Sq equals:

$\frac{\text{q}^3}{2}$
$\text{mnq}$
$\text{q}^3$
$(\text{m}^2+\text{n}^2)$

Answer

$\text{q}^3$

Solution:

As, $\text{S}_\text{n}=\text{n}^2\text{q}$

$\Rightarrow\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]=\text{n}^2\text{q}$

$\Rightarrow2\text{a}+(\text{n}-1)\text{d}=\frac{\text{n}^2\text{q}\times2}{\text{n}}$

$\Rightarrow2\text{a}+(\text{n}-1)\text{d}=2\text{nq}\ .....(1)$

Also, $\text{S}_\text{m}=\text{m}^2\text{q}$

$\Rightarrow\frac{\text{m}}{2}[2\text{a}+(\text{m}-1)\text{d}]=\text{m}^2\text{q}$

$\Rightarrow2\text{a}+(\text{m}-1)\text{d}=\frac{\text{m}^2\text{q}\times2}{\text{m}}$

$\Rightarrow2\text{a}+(\text{m}-1)\text{d}=2\text{mq}\ .....(2)$

Subtracting (1) from (2), we get

$2\text{a}+(\text{n}-1)\text{d}-2\text{a}-(\text{m}-1)\text{d}=2\text{nq}-2\text{mq}$

$\Rightarrow(\text{n}-1-\text{m}+1)\text{d}=2\text{nq}-2\text{mq}$

$\Rightarrow(\text{n}-\text{m})\text{d}=2\text{q}(\text{n}-\text{m})$

$\Rightarrow\text{d}=\frac{2\text{q}(\text{n}-\text{m})}{(\text{n}-\text{m})}$

$\Rightarrow\text{d}=2\text{q}$

Substituting $\text{d}=2\text{q}$ in (2), we get

$2\text{a}+(\text{m}-1)2\text{q}=2\text{mq}$

$\Rightarrow2\text{a}+2\text{mq}-2\text{q}=2\text{mp}$

$\Rightarrow2\text{a}=2\text{q}$

$\Rightarrow\text{a}=\text{q}$

Now,

$\text{S}_\text{q}=\frac{\text{q}}{2}[2\text{a}+(\text{q}-1)\text{d}]$

$=\frac{\text{q}}{2}[2\text{q}+(\text{q}-1)2\text{q}]$

$=\frac{\text{q}}{2}[2\text{q}+2\text{q}^2-2\text{q}]$

$=\frac{\text{q}}{2}[2\text{q}^2]$

$=\text{q}^3$

Hence, the correct alternative is option (c).

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Question 271 Mark

Find the sum of the following arithmetic progression:

1, 3, 5, 7, ... to 12 terms.

Answer

1, 3, 5, 7, ... to 12 terms
$\text{s}_{12}=\frac{12}{2}[2\times1+(12-1)2]$
$=6\times24=144$

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Question 281 Mark

Write the sum of first n even natural numbers.

Answer

We have,
2 + 4 + 6 + ... up to n terms
$\therefore\text{S}_\text{n}=\frac{\text{n}}{2}[2\times2+(\text{n}-1)\times2]$
$=\frac{\text{n}}{2}[4+2\text{n}-2]$
$=\frac{\text{n}}{2}[2\text{n}+2]$
$=\frac{\text{n}}{2}\times2(\text{n}+1)$
$\text{n}(\text{n}+1)$
$\therefore2+4+6+\ ...$ up to n terms = n (n + 1)
Hence, the sum of first n even natural numbers is n (n + 1)

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Question 291 Mark

If the sums of n terms of two arithmetic progressions are in the ratio 2n + 5 : 3n + 4, then write the ratio of their mth terms.

Answer

It is given that
$\frac{\text{S}_\text{n}}{\text{s}_\text{n}}=\frac{2\text{n}+5}{3\text{n}+4}$
Let, $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}_1+(\text{n}-1)\text{d}_1]$
and, $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}_2+(\text{n}-1)]\text{d}_2$
$\therefore\frac{\text{S}_\text{n}}{\text{S}_\text{n}}=\frac{\frac{\text{n}}{2}[2\text{a}_1+(\text{n}-1)\text{d}_2]}{\frac{\text{n}}{2}[2\text{a}_\text{n}+(\text{n}-1)\text{d}_2]}$
$\Rightarrow\frac{2\text{a}_1+(\text{n}-1)\text{d}_1}{2\text{a}_2+(\text{n}-1)\text{d}_2}=\frac{2\text{n}+5}{3\text{n}+4}$
$\Rightarrow\frac{\text{a}_1\Big(\frac{\text{n}-1}{2}\text{}\Big)\text{d}_1}{\text{a}_2+\Big(\frac{\text{n}-1}{2}\Big)\text{d} _2}=\frac{2\text{n}+5}{3\text{n}+4}$
Let, $\frac{\text{n}-1}{2}=\text{m}$
$\Rightarrow\text{n}-1=2\text{m}$
$\Rightarrow\text{n}=2\text{m}-1$
Replacing $\frac{\text{n}-1}{2}$ by m on both side of equation (2), we get
$\frac{\text{a}_1+\text{md}_1}{\text{a}_2\text{md}_2}=\frac{2(2\text{m}-1)+5}{3(2\text{m}-1)+4}$
$=\frac{4\text{m}-2+5}{6\text{m}-3+4}$
$=\frac{4\text{m}+3}{6\text{m}+1}$
$\therefore$ Ratio of m^th terms $=\frac{4\text{m}+3}{6\text{m}+1}$

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Question 301 Mark

If the sums of n terms of two AP.'s are in the ratio (3n + 2) : (2n + 3), then find the ratio of their 12th terms.

Answer

Let the first terms of the two A.P.'s be a and a'; and their common difference be d and d'.
Now,
$\frac{\text{s}_\text{n}}{\text{s}_\text{n}}=\frac{(3\text{n}+2)}{(2\text{n}+3)}$
$\Rightarrow\frac{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]}{\frac{\text{n}}{2}[2\text{a}'+(\text{n}-1)\text{d}]}=\frac{(3\text{n}+2)}{(2\text{n}+3)}$
$\Rightarrow\frac{[2\text{a}+(\text{n}-1)]\text{d}}{[2\text{a}'+(\text{n}-1)\text{d}']}=\frac{(3\text{n}+2)}{(2\text{n}+3)}$
let n = 23
$\Rightarrow\frac{2\text{a}+(23-1)\text{d}}{2\text{a}'+(23-1)\text{d}'}=\frac{3\times23+2}{2\times23+3}$
$\Rightarrow\frac{2\text{a}+22\text{d}}{2\text{a}+22\text{d}}=\frac{69+2}{46+3}$
$\Rightarrow\frac{2(\text{a}+11\text{d)}}{2(\text{a}'11\text{d}')}=\frac{71}{49}$
$\therefore\frac{\text{a}_{12}}{\text{a}_{12}}=\frac{71}{49}$
So, the ratio of there 12th terms is 71 : 49.

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Question 311 Mark

Write the sum of first n odd natural numbers.

Answer

We have,
1 + 3 + 5 + ... up to n terms
$\therefore\text{S}_\text{n}=\frac{\text{n}}{2}[2\times1+(\text{n}-1)\times2]$
$=\frac{\text{n}}{2}[2+2\text{n}-2]$
$=\frac{\text{n}}{2}\times2\text{n}$
$=\text{n}^2$
$\therefore1+3+5+\ ...$ up to n terms = n²
Hence, sum of first n odd natural numbers is n².

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Question 321 Mark

Write the value of n for which nth terms of the A.P.s 3, 10, 17, ... and 63, 65, 67, .... are equal.

Answer

We have,
3 + 10 + 17 + ...
⇒ a_n = 3 + (n - 1)2
and, 63, 65, 67
⇒ a_n =63 + (n - 1)2
It is given that
a_n = a_n
⇒ 3 + (n - 1)7 = 63 + (n - 1)2
⇒ 3 + 7n - 7 = 63 + 2n - 2
⇒ 7n -2n = 61 + 4
⇒ 5n = 65
⇒ n = 13

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Question 331 Mark

If log 2, log (2^x − 1) and log (2^x + 3) are in A.P., write the value of x.

Answer

Since $\log(2^\text{x}-1)$ and $\log(2^\text{x}+3)$ are in A.P, therefore
$\log(2^\text{x}-1)-\log2=\log(2^\text{x}+3)-\log(2^\text{x}-1)$
$2\log(2^\text{x}-1)=\log(2^\text{x}+3)+\log2$
(2^x - 1)² = 2 (2^x + 3)
2^2x - 2 × 2 ^x + 1 = 2.2 ^x + 6
2^2x - 4 × 2 ^x - 5 = 0
2^2x + 2^x - 5 × 2^x - 5 = 0
2^x (2^x + 1) - 5 (2^x + 1) = 0
(2^x + 1)(2^x - 5) = 0
2^x - 5 = 0
2^x = 5
$\text{x}\log2=\log5$
$\text{x}=\frac{\log5}{\log2}$
$=\log_25$

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Question 341 Mark

If $\frac{3+5+7+\ ...\ +\text{Upto n terms}}{5+8+11+\ ...\ \text{Upto 10 terms}}=7,$ then find the value of n.

Answer

We have,
$\frac{3+5+7+\ ...\ +\text{Upto n terms}}{5+8+11+\ ...\ \text{Upto 10 terms}}=7$
$\Rightarrow\frac{\frac{\text{n}}{2}[2\times3+(\text{n}-1)2]}{\frac{10}{2}[2\times5+(10-1)3]}=7$
$\Rightarrow\frac{\text{n}[6+2\text{n}-2]}{10[10+27]}=7$
$\Rightarrow\frac{\text{n}[4+2\text{n}]}{10\times37}=7$
$\Rightarrow4\text{n}+2\text{n}^2=7\times370$
$\Rightarrow2\text{n}+\text{n}^2=7\times185$
$\Rightarrow\text{n}^2+2\text{n}=1295$
$\Rightarrow\text{n}^2+2\text{n}-1295=0$
$\Rightarrow\text{n}^2+37\text{n}-35\text{n}-1295=0$
$\Rightarrow\text{n}(\text{n}+37)-35(\text{n}+37)=0$
$\Rightarrow(\text{n}=37)(\text{n}-35)=0$
$\Rightarrow\text{n}-35=0$
$\Rightarrow\text{n}=35$

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Question 351 Mark

If the sum of n terms of an AP is 2n² + 3n, then write its nth term.

Answer

We have,
S_n = 2n² + 3n
⇒ S_n-1 = 2 (n - 1)² + 3 (n - 1)
= 2 (n² + 1 - 2n) +3n - 3
=2n² + 2 -4n + 3n - 3
= 2n² - n - 1
⇒ S_n-1 = 2n²- n - 1
$\therefore$ T_n = S_n - S_n-1 = 2n² + 3n - (2n² - n - 1)
= 2n² + 3n - 2n² + n + 1
= 4n + 1
Hence, T_n= 4n + 1.

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Question 361 Mark

Write the sum of first n even natural numbers.

Answer

We have,
2 + 4 + 6 + ... up to n terms
$\therefore\text{S}_\text{n}=\frac{\text{n}}{2}[2\times2+(\text{n}-1)\times2]$
$=\frac{\text{n}}{2}[4+2\text{n}-2]$
$=\frac{\text{n}}{2}[2\text{n}+2]$
$=\frac{\text{n}}{2}\times2(\text{n}+1)$
$\text{n}(\text{n}+1)$
$\therefore2+4+6+\ ...$ up to n terms = n (n + 1)
Hence, the sum of first n even natural numbers is n (n + 1)

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Question 371 Mark

Write the common difference of an A.P. whose nth term is xn + y.

Answer

We have,
a_n = xn + y
Putting n =1, we get
a₁ = x + y
Putting n = 2, get
a₂ = 2x + y
$\therefore$ a₂ - a₁ = 2x + y - x - y
= x
⇒ a₂ - a₁ = x
$\therefore$ Common difference = x

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Question 381 Mark

Write the common difference of an A.P. the sum of whose first n terms is $\frac{\text{p}}{2}\text{n}^2+\text{Qn}.$

Answer

We have,
$\text{S}_\text{n}=\frac{\text{p}}{2}\text{n}^2+\text{Qn}$
$\therefore\text{S}_2=\frac{\text{p}}{2}(2)^2+\text{Qx}^2$
$\Rightarrow\text{S}_2=2\text{p}+2\text{Q}$
$\text{S}_3=\frac{\text{p}}{2}(3)^2+3\text{Q}$
$\Rightarrow\text{S}_3=\frac{\text{qp}}{2}+3\text{Q}$
and,
$\text{S}_4=\frac{\text{P}}{2}(4)^2+4\text{Q}$
$\Rightarrow\text{S}_4=8\text{P}+4\text{Q}$
Now,
$\text{T}_4=\text{S}_4-\text{S}_3=8\text{p}+4\text{Q}-\frac{\text{pq}}{2}-3\text{Q}$
$\Rightarrow\text{T}_4=\frac{7\text{P}}{2}+\text{Q}$
and,
$\text{T}_3=\text{S}_3-\text{S}_2=\frac{\text{pq}}{2}+3\text{Q}-2\text{P}-2\text{Q}$
$=\frac{5\text{P}}{2}+\text{Q}$
$\therefore\text{T}_4-\text{T}_3=\frac{7\text{p}}{2}+\text{Q}-\frac{5\text{p}}{2}-\text{Q}$
$=\frac{7\text{P}}{2}-\frac{5\text{P}}{2}$
$=\frac{2\text{P}}{2}$
$=\text{P}$
$\therefore\text{T}_4-\text{T}_3=\text{P}$
$\therefore$ Common difference = P.

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