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

Page No 5.5

Ques.2. Find the indicated terms in each of the following sequences whose nth terms are:
(a) a_n = 5n − 4; a₁₂ and a₁₅
(b)
(c) a_n = n (n −1) (n − 2); a₅ and a₈
(d) a_n = (n − 1) (2 − n) (3 + n); a₁, a₂, a₃
(e) a_n = (−1)n n ; a₃, a₅, a₈
Ans. Here, we are given the n^th term for various sequences. We need to find the indicated terms of the A.P.
(a) a_n = 5n − 4
We need to find a₁₂ and a₁₅
Now, to find a12 term we use n = 12, we get,
a₁₂ = 5(12) - 4
= 60 - 4 = 56
Also, to find a₁₅ term we use n = 15, we get,
a₁₅ = 5(15) − 4
= 75 − 4
= 71
Thus, a₁₂ = 56 and a₁₅ = 71
(b) 
We need to find a₇ and a₈
Now, to find a₇ term we use n = 7, we get

Also, to find a₈ term we use n = 8, we get,

Thus,
(c) a_n = n(n - 1)(n - 2)
We need to find a₅ and a₈
Now, to find a₅ term we use n = 5, we get,
a₅ = 5(5 - 1)(5 - 2)
= 5(4)(3) = 60
Also, to find a₈ term we use n = 8, we get,
a₈ = 8(8 - 1)(8 - 2)
= 8(7)(6) = 336
Thus, a₅ = 60 and a₈ = 336
(d) a_n = (n - 1)(2 - n)(3 + n)
We need to find a₁, a₂ and a₃
Now, to find a₁ term we use n = 1, we get,
a₁ = (1 - 1)(2 - 1)(3 + 1)
= (0)(1)(4) = 0
Also, to find a₂ term we use n = 2, we get,
a₂ = (2 - 1)(2 - 2)(3 + 2)
= (1)(0)(5)
= 0
Similarly, to find a₃ term we use n = 3, we get,
a₃ = (3 - 1)(2 - 3)(3 + 3)
= (2)(-1)(6) = -12
Thus, a₁ = 0, a₂ = 0 and a₃ = -12
(e) a_n = (-1)ⁿn
We need to find a₃, a₅ and a₈
Now, to find a₃ term we use n = 3, we get,
a₃ = (-1)³3
= (-1)3 = -3
Also, to find a₅ term we use n =5, we get,
a₅ = (-1)⁵5
= (-1)5 = -5
Similarly, to find a₈ term we use n = 8, we get,
a₈ = (-1)⁸8
= (1)8 = 8
Thus, a₃ = -3, a₅ = -5 and a₈ = 8.

Ques.3. Find the next five terms of each of the following sequences given by:
(i) a₁ = 1, a_n = a_n−1 + 2, n ≥ 2
(ii) a₁ = a₂ = 2, a_n = a_n−1 − 3, n > 2
(iii) a₁ = −1, a_n = , n ≥ 2
(iv) a₁ = 4, a_n = 4a_n−1 + 3, n > 1.
Ans. In the given problem, we are given the first, second term and the n^th term of an A.P.
We need to find its next five terms
(i) a₁ = 1, a_n = a_n−1 + 2, n ≥ 2
Here, we are given that n ≥ 2
So, the next five terms of this A.P would be a₂, a₃, a₄, a₅ and a₆
Now a₁ = 1 ...(1)
So, to find the a₂ term we use n = 2, we get,
a₂ = a_{2 - 1} + 2
a₂ = a₁ + 2
a₂ = 1 + 2 (Using 1) 
a₂ = 3 … (2) 
For a₃, using n = 3, we get,
a₃ = a_3-1 + 2
a₃ = a₂ + 2
a₃ = 3 + 2(Using 2)
a₃ = 5 ... (3)
For a_4, using n = 4, we get,
a₄ = a_4-1 + 2
a₄ = a₃ + 2
a₄ = 5 + 2(Using 3)
a₄ = 7 ...(4)
For a₅, using n = 5, we get,
a₅ = a_5-1 + 2
a₅ = a₄ + 2
a₅ = 7 + 2 (Using 4)
a₅ = 9 ...(5)
For a₆, using n = 6, we get,
a₆ = a_6-1 + 2
a₆ = a₅ + 2
a₆ = 9 + 2 (Using 5)
a₆ = 11
Therefore, the next five terms, of the given A.P. are a₂ = 3, a₃ = 5, a₄ = 7, a₅ = 9, a₆ = 11
(ii) a₁ = a₂ = 2, a_n = a_n-1 - 3, a > 2
Here, we are given that n > 2
So, the next five terms of this A.P would be a₃, a₄, a₅, a₆ and a₇
Now a₁ = a₂ = 2 ...(1)
So, to find the a₃ term we use n = 3, we get,
a₃ = a_{3 - 1} - 3
a₃ = a₂ - 3
a₃ = 2 - 3 (Using 1)
a₃ = -1 ...(2)
For a₄, using n = 4, we get,
a₄ = a_{4 - 1} - 3
a₄ = a₃ - 3
a₄ = -1 - 3 (Using 2)
a₄ = -4 ...(3)
For a₅, using n = 5, we get,
a₅ = a_{5 - 1} - 3
a₅ = a₄ - 3
a₅ = -4 - 3 (Using 3)
a₅ = - 7 ...(4)
For a₆, using n = 6, we get,
a₆ = a_{6 - 1} - 3
a₆ = a₅ - 3
a₆ = - 7 - 3 (Using 4)
a₆ = -10 ...(5)
For a₇, using n = 7, we get,
a₇ = a_{7 - 1} - 3
a₇ = a₆ - 3
a₇ = - 10 - 3 (Using 5)
a₇ = - 13
Therefore, the next five terms, of the given A.P are a₃ = -1, a₄ = -4, a₅ = -7, a₆ = -10, a₇ = -13
(iii) , ,
Here, we are given that n ≥ 2
So, the next five terms of this A.P would be a₂, a₃, a₄, a5 and a₆
Now a₁ = - 1… (1)
So, to find the a₂ term we use n = 2, we get,

(Using 1)
…… (2)
For a₃, using n = 3, we get,

(Using 2)
 … (3)
For a₄, using n = 4, we get,

(Using 3)
… (4)
For a₅, using n = 5, we get,

(Using 4)
…… (5)
For a₆, using n = 5, we get,

(Using 5)

Therefore, the next five terms, of the given A.P are

(iv) a₁ = 4, an = 4a_n-1 + 3, n > 1
Here, we are given that n > 1.
So, the next five terms of this A.P would be a₂, a₃, a₄, and a₆
Now a₁ = 4… (1)
So, to find the a₂ term we use n = 2, we get,
a₂ = 4a_{2 - 1} + 3
a₂ = 4a₁ + 3
a₂ = 4(4) + 3(Using 1)
a₂ = 19… (2)
For a₃, using n = 3, we get,
a₃ = 4a_{3 - 1} + 3 (Using 2)
a₃ = 4a₂ + 3
a₃ = 4(19) + 3
a₃ = 79 ...(3)
For a_4, using n = 4, we get,
a₄ = 4a_{4 - 1} + 3
a₄ = 4a₃ + 3
a₄ = 4(79) + 3 (Using 3)
a₄ = 319 ... (4)
For a₅, using n = 5, we get,
a₅ = 4a_{5 - 1} + 3
a₅ = 4a₄ + 3
a₅ = 4(319) + 3  (Using 4)
a₅ = 1279 … (5)
For a₆, using n = 6, we get,
a₆ = 4a_6-1 + 3
a₆ = 4a₅ + 3
a₆ = 4(1279) + 3 (Using 5)
a₆ = 5119
Therefore, the next five terms, of the given A.P are
a₂ = 19, a₃ = 79, a₄ = 319, a₅ = 1279, a₆ = 5119

Page No 5.51

Ques.2. Find the sum to n term of the A.P. 5, 2, −1, −4, −7, ...,
Ans. In the given problem, we need to find the sum of the n terms of the given
A.P. “5,2,-1,-4,-7,..." .
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
For the given A.P. (5,2,-1,-4,-7, ...),
Common difference of the A.P. (d) = a₂ - a₁
= 2 - 5 = -3
Number of terms (n) = n
First term for the given A.P. (a) = 5
So, using the formula we get,
S_n = n/2[2(5) + (n - 1) (-3)]
= n/2[10 + (-3n + 3)]
= n/2[10 - 3n + 3]
= n/2[13 - 3n]
Therefore, the sum of first n terms for the given A.P. is n/2[13 - 3n].

Ques.3. Find the sum of n terms of an A.P. whose nth terms is given by an = 5 − 6n.
Ans. Here, we are given an A.P., whose n^th term is given by the following expression, a_n = 5 - 6n
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 = 5 - 6(1)
= 5 - 6 = -1
Now, the last term (l) or the n^th term is given
a_n = 5 - 6n
So, on substituting the values in the formula for the sum of n terms of an A.P., we get,
S_n = (n/2)[(-1) + 5 - 6n]
= (n/2)[4 - 6n]
= (n/2)(2)[2 - 3n]
= (n)(2 - 3n)
Therefore, the sum of the n terms of the given A.P. is (n)(2 - 3n).

Ques.4. Find the sum of last ten terms of the A.P. : 8, 10, 12, 14,...., 126.
Ans. The given A.P is 8, 10, 12, 14,...., 126.
a = 8 and d = 2.
When this A.P is reversed, we get the A.P.
126, 124, 122, 120,....
So, first term becomes 126 and common difference −2.
The sum of first 10 terms of this A.P is as follows:
S₁₀ = 10/2 [2 x 126 + 9(2)]
= 5[234]
= 1170

Ques.5. Find the sum of the first 15 terms of each of the following sequences having nth term as
(i) a_n = 3 + 4n
(ii) b_n = 5 + 2n
(iii) x_n = 6 − n
(iv) y_n = 9 − 5n
Ans.(i) Here, we are given an A.P. whose n^th term is given by the following expression, a_n = 3 + 4n . We need to find the sum of first 15 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 = 3 + 4(1)
= 3 + 4
= 7
Now, the last term (l) or the n^th term is given
l = a_n = 3 + 4n
So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

= (15)(35) = 525
Therefore, the sum of the 15 terms of the given A.P. is S₁₅ = 525.
(ii) Here, we are given an A.P. whose n^th term is given by the following expression
We need b_n = 5 + 2n to find the sum of first 15 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.
b = 5 + 2(1)
= 5 + 2 = 7
Now, the last term (l) or the n^th term is given
l = b_n = 5 + 2n
So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

= (15)(21) = 315
Therefore, the sum of the 15 terms of the given A.P. is S₁₅ = 315.
(iii) Here, we are given an A.P. whose nth term is given by the following expression, x_n = 6 - n. We need to find the sum of first 15 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.
x = 6 - 1 = 5
Now, the last term (l) or the n^th term is given
l = a_n = 6 - n
So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

= (15)(-2) = -30
Therefore, the sum of the 15 terms of the given A.P. is S₁₅ = -30.
(iv) Here, we are given an A.P. whose n^th term is given by the following expression, yn = 9 - 5n. We need to find the sum of first 15 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.
y = 9 - 5(1)
= 9 - 5 = 4
Now, the last term (l) or the n^th term is given
l = a_n = 9 - 5n
So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

= (15)(-31) = -465
Therefore, the sum of the 15 terms of the given A.P. is = S₁₅ = -465.