Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-14) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 5.52

Ques.22. (i) If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of n terms.
(ii) If the sum of first four terms of an A.P. is 40 and that of first 14 terms is 280. Find the sum of its first n terms.
Ans. (i) In the given problem, we need to find the sum of n terms of an A.P. Let us take the first term as a and the common difference as d.
Here, we are given that,
S₇ = 49 ...(1)
S₁₇ = 289 ...(2)
So, as we know the formula for the sum of n terms of an A.P. is given by,
S_n = n/2[2a + (n - 1)d]
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
So, using the formula for n = 7, we get,
S₇ = 7/2[2(a) + (7 - 1)(d)]
49 = (7/2)[2a + (6)(d)] (Using 1)

49 = 7a + 21d
Further simplifying for a, we get,

a = 7 - 3d ...(3)
Also, using the formula for n = 17, we get,

 (Using 2)

289 = 17a + 136d
Further simplifying for a, we get,

a = 17 - 8d ...(4)
Subtracting (3) from (4), we get,
a - a = (17 - 8d)-(7 - 3d)
0 = 17 - 8d - 7 + 3d
0 = 10 - 5d
5d = 10
d = 2
Now, to find a, we substitute the value of d in (3),
a = 7 - 3(2)
a = 7 - 6 = 1
Now, using the formula for the sum of n terms of an A.P., we get,

= n/2[2 + 2n - 2]
= (n/2)(2n) = n²
Therefore, the sum of first n terms for the given A.P. is S_n = n².
(ii) Given:
S₄ = 40
S₁₄ = 280
Now,

⇒ 40 = 2[2a + 3d]
⇒ 2a +3d = 20 ...(1)
S₁₄ = (14/2)[2a + (14 - 1)d]
⇒ 280 = 7[2a + 13d]
⇒ 2a + 13d = 40 ...(2)
Subtracting (1) from (2), we get
10d = 20
⇒ d = 2
Substituting the value of d in (1), we get
a = 7
Therefore, a = 7 and d = 2
Thus,
S_n = (n/2)(2 x 7 + (n - 1)2)
= (n/2)(14 + 2n - 2)
= (n/2)(12 + 2n)
= n(6 + n)
= n² + 6n
Hence,  the sum of its first n terms is n² + 6n.

Ques.23. The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Ans. In the given problem, we have the first and the last term of an A.P. along with the sum of all the terms of A.P. Here, we need to find the number of terms and the common difference of the A.P.
Here,
The first term of the A.P (a) = 5
The last term of the A.P (l) = 45
Sum of all the terms S_n = 400
Let the common difference of the A.P. be d.
So, let us first find the number of the terms (n) using the formula,
400 = (n/2)(5 + 45)
400 = (n/2)(50)
400 = (n)(25)
n = 400/25
n = 16
Now, to find the common difference of the A.P. we use the following formula,
l = a + (n - 1)d
We get,
45 = (5) + (16 - 1)d
45 = 5 + (15)d
45 = 5 + 15d

Further, solving for d,
d = 40/15
d = 8/3
Therefore, the number of terms is n = 16 and the common difference of the A.P is d = 8/3.

Ques.24. In an A.P. the first term is 8, nth term is 33 and the sum to first n terms is 123. Find n and d, the common differences.
Ans. In the given problem, we have the first and the nth term of an A.P. along with the sum of the n terms of A.P. Here, we need to find the number of terms and the common difference of the A.P.
Here,
The first term of the A.P (a) = 8
The nth term of the A.P (l) = 33
Sum of all the terms S_n = 123
Let the common difference of the A.P. be d.
So, let us first find the number of the terms (n) using the formula,


n = 246/41 = 6
Now, to find the common difference of the A.P. we use the following formula,
l = a + (n - 1)d
We get,
33 = 8 + (6 - 1)d
33 = 8 + (5)d

Further, solving for d,
d = 25/5 = 5
Therefore, the number of terms is n = 6 and the common difference of the A.P. d = 5.

Ques.25. In an A.P., the first term is 22, nth term is −11 and the sum to first n terms is 66. Find n and d, the common difference
Ans. In the given problem, we have the first and the nth term of an A.P. along with the sum of the n terms of A.P. Here, we need to find the number of terms and the common difference of the A.P.
Here,
The first term of the A.P (a) = 22
The nth term of the A.P (l) = −11
Sum of all the terms S_n = 66
Let the common difference of the A.P. be d.
So, let us first find the number of the terms (n) using the formula,

(66)(2) = (n)(11)
Further, solving for n

n = (6)(2) = 12
Now, to find the common difference of the A.P. we use the following formula,
l = a + (n - 1)d
We get,
-11 = 22 + (12 - 1)d
-11 = 22 + (11)d

Further, solving for d,
d = (-33)/11
d = -3
Therefore, the number of terms is n = 12 and the common difference of the A.P. d = -3.

Ques.26.The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find its common difference.
Ans. Let a be the first term and d be the common difference.
We know that, sum of first n terms = S_n = (n/2)[2a + (n - 1)d]
Also, n^th term = a_n = a + (n − 1)d
According to the question,
a = 7, a_n = 49 and S_n = 420
Now,
a_n = a + (n − 1)d
⇒ 49 = 7 + (n − 1)d
⇒ 42 = nd − d
⇒ nd − d = 42 ...(1)
Also,

⇒ 840 = n[14 + 42] [From (1)]
⇒ 56n = 840
⇒ n = 15 ...(2)
On substituting (2) in (1), we get
nd − d = 42
⇒ (15 − 1)d = 42
⇒ 14d = 42
⇒ d = 3
Thus, common difference of the given A.P. is 3.

Ques.27. The first and the last terms of an A.P. are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference.
Ans. Let a be the first term and d be the common difference.
We know that, sum of first n terms = S_n = n/2[2a + (n - 1)d]
Also, n^th term = a_n = a + (n − 1)d
According to the question,
a = 5, a_n = 45 and S_n = 400
Now,
a_n = a + (n − 1)d
⇒ 45 = 5 + (n − 1)d
⇒ 40 = nd − d
⇒ nd − d = 40 ...(1)
Also,
S_n = n/2[2 × 5 + (n − 1)d]
⇒ 400 = n/2[10 + nd − d]
⇒ 800 = n[10 + 40] [From (1)]
⇒ 50n = 800
⇒ n = 16 ....(2)
On substituting (2) in (1), we get
nd − d = 40
⇒ (16 − 1)d = 40
⇒ 15d = 40
⇒ d = 8/3
Thus, common difference of the given A.P. is 8/3.

Ques.28. The sum of first 9 terms of an A.P. is 162. The ratio of its 6^th term to its 13^th term is 1 : 2. Find the first and 15^th term of the A.P.
Disclaimer: There is a misprint in the question, 'the sum of 9 terms' should be witten instead of 'the sum of q terms'.
Ans. Let a be the first term and d be the common difference.
We know that, sum of first n terms = S_n = (n/2)[2a + (n − 1)d]
Also, n^th term = a_n = a + (n − 1)d
According to the question,
S_q = 162 and
Now,

⇒ 2a + 10d = a + 12d
⇒ 2a − a = 12d −  10d
⇒ a = 2d .....(1)
Also,
S₉ = 9/2[2a + (9 − 1)d]
⇒ 162 = 9/2[2(2d) + 8d] [From (1)]
⇒ 18 = 1/2[12d]
⇒ 18 = 6d
⇒ d = 3
⇒ a = 2 × 3 [From (1)]
⇒ a = 6
Thus, the first term of the A.P. is 6.
Now,
a₁₅ = 6 + (15 − 1)3
= 6 + 42 = 48
Thus, 15^th term of the A.P. is 48.

Ques.29. If the 10^th term of an A.P. is 21 and the sum of its first 10 terms is 120, find its n^th term.
Ans. Let a be the first term and d be the common difference.
We know that, sum of first n terms = S_n = n/2[2a + (n − 1)d]
and n^th term = a_n = a + (n − 1)d
Now,
S₁₀ = 10/2[2a + (10 − 1)d]
⇒ 120 = 5(2a + 9d)
⇒ 24 = 2a + 9d
⇒ 2a + 9d = 24 ....(1)
Also,
a₁₀ = a + (10 − 1)d
⇒ 21 = a + 9d
⇒ 2a + 18d = 42 ....(2)
Subtracting (1) from (2), we get
18d − 9d = 42 − 24
⇒ 9d = 18
⇒ d = 2
⇒ 2a = 24 − 9d [From (1)]
⇒ 2a = 24 − 9 × 2
⇒ 2a = 24 − 18
⇒ 2a = 6 = 3
Also,
a_n = a + (n − 1)d
= 3 + (n − 1)2
= 3 + 2n − 2
= 1 + 2n
Thus, n^th term of this A.P. is 1 + 2n.

Ques.30. The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28^th term of this A.P.
Ans. Let a be the first term and d be the common difference.
We know that, sum of first n terms = S_n = n/2[2a + (n − 1)d]
It is given that sum of the first 7 terms of an A.P. is 63.
And sum of next 7 terms is 161.
∴ Sum of first 14 terms = Sum of first 7 terms + sum of next 7 terms
= 63 + 161 = 224
Now,
S₇ = 7/2[2a + (7 − 1)d]
⇒ 63 = 7/2(2a + 6d)
⇒ 18 = 2a + 6d
⇒ 2a + 6d = 18 ....(1)
Also,
S₁₄ = 14/2[2a + (14 − 1)d]
⇒ 224 = 7(2a + 13d)
⇒ 32 = 2a + 13d
⇒ 2a + 13d =  32 ....(2)
On subtracting (1) from (2), we get
13d − 6d = 32 − 18
⇒  7d = 14 = 2
⇒ 2a = 18 − 6d [From (1)]
⇒ 2a = 18 − 6 × 2
⇒ 2a = 18 − 12
⇒ 2a = 6 = 3
Also, n^th term = a_n = a + (n − 1)d
⇒ a₂₈ = 3 + (28 − 1)2
= 3 + 27 × 2 = 57
Thus, 28^th term of this A.P. is 57.

Ques.31. The sum of first seven terms of an A.P. is 182. If its 4^th and the 17^th terms are in the ratio 1 : 5, find the A.P.
Ans. Let a be the first term and d be the common difference.
We know that, sum of first n terms = S_n = n/2[2a + (n − 1)d]
According to the question,
S₇ = 182
⇒ 7/2[2a + (7 − 1)d] =182
⇒ 1/2(2a + 6d) = 26
⇒ a + 3d = 26
⇒ a = 26 − 3d ....(1)
Also,

⇒ 5(a + 3d) = a + 16d
⇒ 5a + 15d = a + 16d
⇒ 5a − a = 16d − 15d
⇒ 4a = d ....(2)
On substituting (2) in (1), we get
a = 26 − 3(4a)
⇒ a = 26 − 12a
⇒ 12a + a = 26
⇒ 13a = 26
⇒ a = 2
⇒ d = 4 × 2 [From (2)]
⇒ d = 8
Thus, the A.P. is 2, 10, 18, 26, ..... .

Ques.32. The nth term of an A.P is given by (−4n + 15), Find the sum of first 20 terms of this A.P.
Ans. a_n = −4n + 15
⇒ a₁ = −4 + 15 = 11
Also, a₂ = −8 + 15 = 7
Common difference, d = a₂ − a₁ = 7 − 11 = −4
Now,
S₂₀ = 20/2[2 × 11 + (20 − 1)(−4)]
= 10(22 − 76) = -540