Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-15) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 5.52

Ques.33. In an A.P., the sum of first ten terms is −150 and the sum of its next ten terms is −550. Find the A.P.
Ans. Here, we are given S₁₀ = -150 and sum of the next ten terms is −550.
Let us take the first term of the A.P. as a and the common difference as d.
So, let us first find a₁₀. For the sum of first 10 terms of this A.P,
First term = a
Last term = a₁₀
So, we know,
a_n = a + (n - 1)d
For the 10^th term (n = 10),
a₁₀ = a + (10 - 1)d
= a + 9d
So, here we can find the sum of the n terms of the given A.P., using the formula, S_n = (n/2)(a + l)
Where, a = the first term
l = the last term
So, for the given A.P,
S₁₀ = (10/2)(a + a + 9d)
-150 = 5(2a + 9d)
-150 = 10a + 45d
 ...(1)
Similarly, for the sum of next 10 terms (S₁₀),
First term = a₁₁
Last term = a₂₀
For the 11^th term (n = 11),
a₁₁ = a + (11 - 1)d
= a + 10d
For the 20^th term (n = 20),
a₂₀ = a + (20 - 1)d
= a + 19d
So, for the given A.P,

-550 = 5(2a + 29d)
-550 = 10a + 145d
 ...(2)
Now, subtracting (1) from (2),

0 = -400 - 100d
100d = -400 = -4
Substituting the value of d in (1)

= 30/10 = 3
So, the A.P. is 3, -1, -5, -9, ... with a = 3, d = -4.

Ques.34. Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25^th term.
Ans. First term, a = 10
Sum of first 14 terms, S₁₄ = 1505
⇒ 14/2[2 × 10 + (14 − 1)d] = 1505
⇒ 7 × (20 − 13d) = 1505
⇒ 20 − 13d = 1505/7 = 215
⇒ 13d = −195
⇒ d = −15
Now,
a₂₅ = 10 + 24(−15) = −350

Ques.35. In an A.P., the first term is 2, the last term is 29 and the sum of the terms is 155. Find the common difference of the A.P.
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 common difference of the A.P.
Here,
The first term of the A.P (a) = 2
The last term of the A.P (l) = 29
Sum of all the terms (S_n) = 155
Let the common difference of the A.P. be d.
So, let us first find the number of the terms (n) using the formula,
155 = (n/2)(2 + 29)
155 = (n/2)(31)
155(2) = (n)(31)
n = 310/31 = 10
Now, to find the common difference of the A.P. we use the following formula,
l = a + (n - 1)d
We get,
29 = 2 + (10 - 1)d
29 = 2 + (9)d
29 - 2 = 9d
d = 27/9
d = 3
Therefore, the common difference of the A.P. is d = 3.

Page No 5.53

Ques.36. The first and the last term of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Ans. In the given problem, we have the first and the last term of an A.P. along with the common difference of the A.P. Here, we need to find the number of terms of the A.P. and the sum of all the terms.
Here,
The first term of the A.P (a) = 17
The last term of the A.P (l) = 350
The common difference of the A.P. = 9
Let the number of terms be n.
So, as we know that,
l = a + (n - 1)d
We get,
350 = 17 + (n - 1)9
350 = 17 + 9n - 9
350 = 8 + 9n
350 - 8 = 9n
Further solving this,
n = 342/9 = 38
Using the above values in the formula,
S_n = (n/2)(a + l)
= (38/2)(17 + 350)
= (19)(367) = 6973
Therefore, the number of terms is n = 38 and the sum S_n = 6973.

Ques.37. Find the number of terms of the A.P. −12, −9, −6, ..., 21. If 1 is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained.
Ans. First term, a₁ = −12
Common difference, d = a₂ − a₁ = −9 − (−12) = 3
a_n = 21
⇒ a + (n − 1)d = 21
⇒ −12 + (n − 1) × 3 = 21
⇒ 3n = 36
⇒ n = 12
Therefore, number of terms in the given A.P. is 12.
Now, when 1 is added to each of the 12 terms, the sum will increase by 12.
So, the sum of all terms of the A.P. thus obtained
= S₁₂ + 12
= 12/2 [2(−12) + 11(3)] + 12
= 6 × (9) + 12 = 66

Ques.38. The sum of the first n terms of an A.P. is 3n² + 6n. Find the n^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 n terms of an A.P. is 3n² + 6n.
∴ First term = a =  S₁ = 3(1)² + 6(1) = 9.
Sum of first two terms = S₂ = 3(2)² + 6(2) = 24.
∴ Second term = S₂ − S₁ = 24 − 9 = 15.
∴ Common difference = d = Second term − First term
= 15 − 9 = 6
Also, n^th term = a_n = a + (n − 1)d
⇒ a_n = 9 + (n − 1)6
⇒ a_n = 9 + 6n − 6
⇒ a_n = 3 + 6n
Thus, n^th term of this A.P. is 3 + 6n.

Ques.39. The sum of first n terms of an A.P. is 5n − n². Find the n^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 n terms of an A.P. is 5n − n².
∴ First term = a =  S₁ = 5(1) − (1)² = 4.
Sum of first two terms = S₂ = 5(2) − (2)² = 6.
∴ Second term = S₂ − S₁ = 6 − 4 = 2.
∴ Common difference = d = Second term − First term
= 2 − 4 = −2
Also, n^th term = an = a + (n − 1)d
⇒ a_n = 4 + (n − 1)(−2)
⇒ a_n = 4 − 2n + 2
⇒ a_n = 6 − 2n
Thus, n^th term of this A.P. is 6 − 2n.

Ques.40. The sum of the first n terms of an A.P. is 4n² + 2n. Find the n^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 n terms of an A.P. is 4n² + 2n.
∴ First term = a =  S₁ = 4(1)² + 2(1) = 6.
Sum of first two terms = S₂ = 4(2)² + 2(2) = 20.
∴ Second term = S₂ − S₁ = 20 − 6 = 14.
∴ Common difference = d = Second term − First term
= 14 − 6 = 8
Also, n^th term = a_n = a + (n − 1)d
⇒ a_n = 6 + (n − 1)(8)
⇒ a_n = 6 + 8n − 8
⇒ a_n = 8n − 2
Thus, n^th term of this A.P. is 8n − 2.

Ques.41. The sum of first n terms of an A.P. is 3n² + 4n. Find the 25th term of this A.P.
Ans. S_n = 3n² + 4n
We know
a_n = S_n − S_n−1
∴ a_n = 3n² + 4n − 3(n−1)² − 4(n−1)
⇒ a_n = 6n + 1
a₂₅ = 6(25) + 1 = 151

Ques.42. The sum of first n terms of an A.P is 5n² + 3n. If its mth terms is 168, find the value of m. Also, find the 20th term of this A.P.
Ans. S_n = 5n² + 3n
We know
a_n = S_n − S_n−1
∴ a_n = 5n² + 3n − 5(n − 1)² − 3(n−1)
a_n = 10n − 2
Now,
a_m = 168
⇒ 10m − 2 = 168
⇒ 10m = 170
⇒ m = 17
a₂₀ = 10(20) − 2 = 198

Ques.43. The sum of first q terms of an A.P. is 63q − 3q². If its pth term is −60,  find the value of p. Also, find the 11th term of this A.P.
Ans. S_q = 63q − 3q²
We know
a_q = S_q − S_q−1
∴ aq = 63q − 3q² − 63(q − 1) + 3(q − 1)²
a_q = 66 − 6q
Now, a_p = −60
⇒ 66 − 6p = −60
⇒ 126 = 6p
⇒ p = 21
a₁₁ = 66 − 6 × 11 = 0

Ques.44. The sum of first m terms of an A.P. is 4m² − m. If its nth term is 107. find the value of n. Also, find the 21st term of this A.P.
Ans. S_m = 4m² − m
We know
a_m = S_m − S_m−1
∴ a_m = 4m² − m − 4(m − 1)² + (m − 1)