Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-11) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 5.51

Ques.6. Find the sum of first 20 terms of the sequence whose nth term is a_n = An + B.
Ans. Here, we are given an A.P. whose n^th term is given by the following expression a_n = An + B. We need to find the sum of first 20 terms.
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,
The first term (a) will be calculated using n = 1in the given equation for n^th term of A.P.
a = A(1) + B
= A + B
Now, the last term (l) or the n^th term is given
l = a_n = An + B
So, on substituting the values in the formula for the sum of n terms of an A.P., we get,
S₂₀ = (20/2)[(A + B) + A(20) + B]
= 10[21A + 2B]
= 210A + 20B
Therefore, the sum of the first 20 terms of the given A.P. is S₂₀ = 210A + 20B.

Ques.7. Find the sum of the first 25 terms of an A.P. whose nth term is given by a_n = 2 − 3n.
Ans. Here, we are given an A.P. whose nth term is given by the following expression, a_n = 2 - 3n. We need to find the sum of first 25 terms.
So, here we can find the sum of the n terms of the given A.P., using the formula, Sn = (n/2)(a+l)
Where, a = the first term
l = the last term
So, for the given A.P,
The first term (a) will be calculated using n = 1 in the given equation for n^th term of A.P.
a = 2 - 3(1)
= 2 - 3 = -1
Now, the last term (l) or the n^th term is given
l = a_n = 2 - 3n
So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

=(25/2)(-74)
= (25)(-37) = -925
Therefore, the sum of the 25 terms of the given A.P. is S₂₅ = -925.

Ques.8. Find the sum of the first 25 terms of an A.P. whose nth term is given by a_n = 7 − 3n.
Ans. Here, we are given an A.P. whose n^th term is given by the following expression, a_n = 7 - 3n. We need to find the sum of first 25 terms.
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,
The first term (a) will be calculated using n = 1 in the given equation for n^th term of A.P.
a = 7 - 3(1)
= 7 - 3 = 4
Now, the last term (l) or the n^th term is given
l = a_n = 7 - 3n
So, on substituting the values in the formula for the sum of n terms of an A.P., we get,
 
= (25/2)(-64)
= (25)(-32) = -800
Therefore, the sum of the 25 terms of the given A.P. is S_n = -800.

Ques.9. If the sum of a certain number of terms starting from first term of an A.P. is 25, 22, 19, ..., is 116. Find the last term.
Ans. In the given problem, we have the sum of the certain number of terms of an A.P. and we need to find the last term for that A.P.
So here, let us first find the number of terms whose sum is 116. For that, we will use the formula,
S_n = (n/2)(a+l)
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
So for the given A.P (25, 22, 19,...)
The first term (a) = 25
The sum of n terms S_n = 116
Common difference of the A.P. (d) = a₂ - a₁
= 22 - 25 = -3
So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

(116)(2) = 53n - 3n²
So, we get the following quadratic equation,
3n² - 53n + 232 = 0
On solving by splitting the middle term, we get,
3n² - 24n - 29n + 232 = 0
3n(n − 8) − 29(n − 8) = 0
(3n − 29)(n − 8) = 0
Further,
3n - 29 = 0
n = 29/3
Also,
n - 8 = 0
n = 8
Now, since n cannot be a fraction, so the number of terms is 8.
So, the term is a8
a₈ = a₁ + 7d
= 25 + 7(-3)
= 25 - 21 = 4
Therefore, the last term of the given A.P. such that the sum of the terms is 116 is 4.

Ques.10. (i) How many terms of the sequence 18, 16, 14, ... should be taken so that their sum is zero?
(ii) How many terms are there in the A.P. whose first and fifth terms are −14 and 2 respectively and the sum of the terms is 40?
(iii) How many terms of the A.P. 9, 17, 25,... must be taken so that their sum is 636?
(iv) How many terms of the A.P. 63, 60, 57, ... must be taken so that their sum is 693?
(v) How many terms of the A.P 27, 24, 21... should be taken so that their sum is zero?
(vi) How many terms of the A.P 45, 39, 33... must be taken so that their sum is 180?
Explain the double answer.
Ans. In the given problem, we have the sum of the certain number of terms of an A.P. and we need to find the number of terms.
(i) A.P. is 18,16,14,...
So here, let us find the number of terms whose sum is 0. For that, we will use the formula,
S_n = (n/2)(a+l)
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
The first term (a) = 18
The sum of n terms (S_n) = 0
Common difference of the A.P. (d) = a₂ - a₁
= 16 - 18 = -2
So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

0 = (n/2)[38 - 2n]
Further,
n/2 = 0
n = 0
Or,
38 - 2n = 0
n = -38/-2 = 19
Since, the number of terms cannot be zero, the number of terms (n) is n = 19.
(ii) Here, let us take the common difference as d.
So, we are given,
First term (a₁) = −14
Fifth term (a₅) = 2
Sum of terms (s_n) = 40
Now,
a₅ = a₁ + 4d
2 = -14 + 4d
2 + 14 = 4d
d = 16/4 = 4
Further, let us find the number of terms whose sum is 40. For that, we will use the formula,
S_n = (n/2)(a+l)
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
The first term (a₁) = −14
The sum of n terms (S_n) = 40
Common difference of the A.P. (d) = 4
So, on substituting the values in the formula for the sum of n terms of an A.P., we get,
40 = n/2[2(-14) + (n - 1)(4)
40 = (n/2)[-28 + (4n - 4)]
40 = (n/2)[-32 + 4n]
40(2) = -32n + 4n²
So, we get the following quadratic equation,
4n² - 32n - 80 = 0
n² - 8n + 20 = 0
On solving by splitting the middle term, we get,
n² - 10n + 2n + 20 = 0
n(n - 10)+ 2(n - 10) = 0
(n + 2)(n - 10) = 0
Further,
n + 2 = 0
n = -2
Or,
n - 10 = 0
n = 10
Since the number of terms cannot be negative. Therefore, the number of terms (n) is n = 10
(iii) A.P. is 9,17,25,...
So here, let us find the number of terms whose sum is 636. For that, we will use the formula,
S_n = (n/2)(a+l)
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
The first term (a) = 9
The sum of n terms (S_n) = 636
Common difference of the A.P. (d) = a₂ - a₁ 
= 17 - 9 = 18
So, on substituting the values in the formula for the sum of n terms of an A.P., we get,
636 = n/2[2(9) + (n - 1)(8)]
636 = (n/2)[18 + (8n - 8)]
636 = (n/2)[10 + 8n]
636(2) = 10n + 8n²
So, we get the following quadratic equation,
8n² + 10n - 1272 = 0
4n² + 5n - 636 = 0
On solving by splitting the middle term, we get,
4n² - 48n + 53n - 636 = 0
4n(n - 12)-53(n - 12) = 0
(4n - 53)(n - 12) = 0
Further,
4n - 53 = 0
n = 53/4
Or,
n - 12 = 0
n = 12
Since, the number of terms cannot be a fraction, the number of terms (n) is 12.
(iv) A.P. is 63,60,57,...
So here, let us find the number of terms whose sum is 693. For that, we will use the formula,
S_n = (n/2)(a+l)
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
The first term (a) = 63
The sum of n terms (S_n) = 693
Common difference of the A.P. (d) = a₂ - a₁
= 60 - 63 = -3
So, on substituting the values in the formula for the sum of n terms of an A.P., we get,
693 = n/2[2(63) + (n - 1)(-3)]
693 = n/2[126 + (-3n + 3)]
693 = (n/2)[129  - 3n)
693(2) = 129n - 3n²
So, we get the following quadratic equation,
3n² - 129n + 1386 = 0
n² - 43n + 462 = 0
On solving by splitting the middle term, we get,
n² - 22n - 21n + 462 = 0
n(n - 22)-21(n - 22) = 0
(n - 22)(n - 21) = 0
Further,
n - 22 = 0
n = 22
Or,
n - 21 = 0
n = 21
Here, 22^nd term will be
a₂₂ = a₁ + 21d
= 63 + 21(-3)
= 63 - 63 = 0
So, the sum of 22 as well as 21 terms is 693. Therefore, the number of terms (n) is 21 or 22.
(v) The given AP is 27, 24, 21, ...
First term of the AP = 27
Common difference = 24 − 27 = −3
Let the sum of the first x terms of the AP be 0.
Sum of first x terms = x/2[2×27+(x−1)(−3)]=0
⇒x/2[54 + (−3x + 3)]=0
⇒x(54 − 3x + 3)=0
⇒x(57 − 3x)=0
Now, either x = 0 or 57 − 3x = 0.
Since the number of terms cannot be 0, x ≠ 0.
∴ 57 − 3x = 0
⇒ 57 = 3x
⇒ x = 19
Thus, the sum of the first 19 terms of the AP is 0.
(v) 27, 24, 21...
Given:
a = 27
d = 24 − 27 = -3
S_n = 0
Now,
S_n = n/2[2a + (n − 1)d]
⇒ 0 = n/2[2(27) + (n − 1)(−3)]
⇒ 0 = n[2(27) + (n − 1)(−3)]
⇒ n[54 − 3n + 3] = 0
⇒ n(57 − 3n) = 0
⇒ n = 0 or 3n = 57
But n is a natural number, ∴ n ≠ 0
⇒ n = 19
Hence, 19  terms of the A.P 27, 24, 21... should be taken so that their sum is zero.

Ques.11. Find the sum of the first
(i) 11 terms of the A.P.2, 6, 10. 14
(ii) 13 terms of the A.P. −6, 0, 6, 12, ...
(iii) 51 terms of the A.P. : whose second term is 2 and fourth term is 8.
Ans. In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of n terms of an A.P.,
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
(i) 2,6,10,14,... To 11 terms.
Common difference of the A.P. (d) = a₂ - a₁
= 6 - 2 = 4
Number of terms (n) = 11
First term for the given A.P. (a) = 2
So, using the formula we get,
S_n = 11/2[2(2) + (11 - 1)(4)]
=(11/2)[4 + (10)(4)]
= (11/2)[4 + 40]
= (11/2)[44]
= 242
Therefore, the sum of first 11 terms for the given A.P. is 242.
(ii) -6,0,6,12,... To 13 terms.
Common difference of the A.P. (d) = a₂ - a₁
= 0 - (-6) = 6
Number of terms (n) = 13
First term for the given A.P. (a) = −6
So, using the formula we get,

= (13/2)[60] = 390
Therefore, the sum of first 13 terms for the given A.P. is 390.
(iii) 51 terms of an A.P whose a₂ = 2 and a₄ = 8
Now,
a₂ = a + d
2 = a + d ...(1)
Also,
a₄ = a + 3d
8 = a + 3d ...(2)
Subtracting (1) from (2), we get
2d = 6
d = 3
Further substituting d = 3 in (1), we get
2 = a + 3
a = -1
Number of terms (n) = 51
First term for the given A.P. (a) = −1
So, using the formula we get,
S_n = 51/2[2(-1) + (51 - 1)(3)]
= (51/2)[-2 + (50)(3)]
= (51/2)[-2 + 150]
= (51/2)[148]
= 3774
Therefore, the sum of first 51 terms for the given A.P. is 3774.