Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-2) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 5.11

Ques.6. Find the common difference of the A.P. and write the next two terms:
(i) 51, 59, 67, 75, ...
(ii) 75, 67, 59, 51, ...
(iii) 1.8, 2.0, 2.2, 2.4, ...
(iv) 0,1/4,1/2,3/4, ...
(v) 119, 136, 153, 170, ...
Ans. In this problem, we are given different A.P. and we need to find the common difference of the A.P., along with the next two terms.
(i) 51, 59, 67, 75, ...
Here,
a₁ = 51
a₂ = 59
So, common difference of the A.P. (d) = a₂ - a₁
= 59 - 51
= 8
Also, we need to find the next two terms of A.P., which means we have to find the 5^th and 6^th term.
So, for fifth term,
a₅ = a₁ + 4d
= 51 + 4(8)
= 51 + 32
= 83
Similarly, we find the sixth term,
a₆ = a₁ + 5d
= 51 + 5(8)
= 51 + 40
= 91
Therefore, the common difference is d = 8 and the next two terms of the A.P. are a₅ = 83, a₆ = 91.
(ii) 75, 67, 59, 51, ...
Here,
a₁ = 75
a₂ = 67
So, common difference of the A.P. (d) = a₂ - a₁
= 67 - 75
= -8
Also, we need to find the next two terms of A.P., which means we have to find the 5^th and 6^th term.
So, for fifth term,
a₅ = a₁ + 4d
= 75 + 4(-8)
= 75 - 32
= 43
Similarly, we find the sixth term,
a₆ = a₁ + 5d
= 75 + 5(-8)
= 75 - 40
= 35
Therefore, the common difference is d = -8 and the next two terms of the A.P. are a₅ = 43, a₆ = 35.
(iii) 1.8, 2.0, 2.2, 2.4, ...
Here,
a₁ = 1.8
a₂ = 2.0
So, common difference of the A.P. (d) = a₂ - a₁
= 2.0 - 1.8
= 0.2
Also, we need to find the next two terms of A.P., which means we have to find the 5^th and 6^th term.
So, for fifth term,
a₅ = a₁ + 4d
= 1.8 + 4(0.2)
= 1.8 + 0.8
= 2.6
Similarly, we find the sixth term,
a₆ = a₁ + 5d
= 1.8 + 5(0.2)
= 1.8 + 1
= 2.8
Therefore, the common difference is d = 0.2 and the next two terms of the A.P. are a₅ = 2.6, a₆ = 2.8.
(iv) 0,1/4,1/2,3/4, ...
Here,
a₁ = 0
a₂ = 1/4
So, common difference of the A.P. (d) = a₂ - a₁
= 1/4 - 0
= 1/4
Also, we need to find the next two terms of A.P., which means we have to find the 5^th and 6^th term.
So, for fifth term,
a₅ = a₁ + 4d
= 0 + 4(1/4)
= 1
Similarly, we find the sixth term,
a₆ = a₁ + 5d
= 0 + 5(1/4)
= 5/4
Therefore, the common difference is d = 1/4 and the next two terms of the A.P. are a₅ = 1, a₆ = 5/4.
(v) 119, 136, 153, 170, ...
Here,
a₁ = 119
a₂ = 136
So, common difference of the A.P. (d) = a₂ - a₁
= 136 - 119
= 17
Also, we need to find the next two terms of A.P., which means we have to find the 5^th and 6^th term.
So, for fifth term,
a₅ = a₁ + 4d
= 119 + 4(17)
= 119 + 68
= 187
Similarly, we find the sixth term,
a₆ = a₁ + 5d
= 119 + 5(17)
= 119 + 85
= 204
Therefore, the common difference is d = 17 and the next two terms of the A.P. are a₅ = 187, a₆ = 204.

Ques.7. Prove that no matter what the real numbers a and b are, the sequence with nth term a + nb is always an A.P. What is the common difference?
Ans. In the given problem, we are given the sequence with the n^th term (a_n) as a + nb where a and b are real numbers.
We need to show that this sequence is an A.P and then find its common difference (d)
Here,
a_n = a + nb
Now, to show that it is an A.P, we will find its few terms by substituting n = 1, 2, 3
So,
Substituting n = 1, we get
a₁ = a + (1)b
a₁ = a + b
Substituting n = 2, we get
a₂ = a + (2)b
a₂ = a + 2b
Substituting n = 3, we get
a₃ = a + 3(b)
a₃ = a + 3b
Further, for the given to sequence to be an A.P,
Common difference (d) = a₂ - a₁ = a₃ - a₂
Here,
a₂ - a₁ = a + 2b - a - b
= b
Also,
a₃ - a₂ = a + 3b - a - 2b
= b
Since a₂ - a₁ = a₃ - a₂
Hence, the given sequence is an A.P and its common difference is d = b.

Page No 5.24

Ques.1. Find:
(i) 10^th term of the A.P. 1, 4, 7, 10, ...
(ii) 18^th term of the A.P. √2, 3√2, 5√2...
(iii) n^th term of the A.P. 13, 8, 3, −2, ...
(iv) 10^th term of the A.P. −40, −15, 10, 35, ...
(v) 8^th term of the A.P. 117, 104, 91, 78, ...
(vi) 11^th term of the A.P. 10.0, 10.5, 11.0, 11.5, ...
(vii) 9^th term of the A.P. 3/4,5/4,7/4,9/4, ...
Ans. In this problem, we are given different A.P. and we need to find the required term of that A.P.
(i) 10^th term of the A.P. 1, 4, 7, 10, ...
Here,
First term (a) = 1
Common difference of the A.P. (d) = 4 - 1 = 3
Now, as we know,
a_n = a + (n - 1)d
So, for 10^th term,
a₁₀  = a + (10 - 1)d
= 1 + 9(3)
= 1 + 27
= 28
Therefore, the 10^th term of the given A.P. is a₁₀ = 28.
(ii) 18^th term of the A.P. √2, 3√2, 5√2...
Here,
First term (a) = √1
Common difference of the A.P. (d) = 3√2-√2
=2√2
Now, as we know,
a_n = a + (n - 1)d
So, for 18^th term,
a₁₈ = a + (18 - 1)d
= √2 + (17)2√2
= √2 + 34√2
= 35√2
Therefore, the 18^th term of the given A.P. is a₁₈ = 35√2.
(iii) n^th term of the A.P. 13, 8, 3, −2, ...
Here,
First term (a) = 13
Common difference of the A.P. (d) = 8 - 13
= -5
Now, as we know,
a_n = a + (n - 1)d
So, for n^th term,
a_n = a + (n - 1)d
= 13 + (n - 1)(-5)
= 13 + (-5n + 5)
= 13 - 5n + 5
= 18 - 5n
Therefore, the n^th term of the given A.P. is a_n = 18 - 5n.
(iv) 10^th term of the A.P. −40, −15, 10, 35, ...
Here,
First term (a) = -40
Common difference of the A.P. (d) = -15 - (-40)
= -15 + 40
= 25
Now, as we know,
a_n = a + (n - 1)d
So, for 10^th term,
a₁₀ = a + (10 - 1)d
= -40 + (9)25
= -40 + 225
= 185
Therefore, the 10^th term of the given A.P. is a₁₀ = 185.
(v) 8^th term of the A.P. 117, 104, 91, 78, ...
Here,
First term (a) = 117
Common difference of the A.P. (d) = 104 - 117
= - 13
Now, as we know,
a_n = a + (n - 1)d
So, for 8^th term,
a₈ = a + (8 - 1)d
= 117 + (7)(-13)
= 117 - 91
= 26
Therefore, the 8^th term of the given A.P. is a₈ = 26.
(vi) 11^th term of the A.P. 10.0, 10.5, 11.0, 11.5, ...
Here,
First term (a) = 10.0
Common difference of the A.P. (d) = 10.5 - 10.0
= 0.5

Now, as we know,
a_n = a + (n - 1)d
So, for 11^th term,
a₁₁ = a + (11 - 1)d
= 10.0 + (10)(0.5)
= 10.0 - 5.0
= 15.0
Therefore, the 11^th term of the given A.P. is a₁₁ = 15.0.
(vii) 9^th term of the A.P. 3/4,5/4,7/4,9/4, ...
Here,
First term (a) = 3/4
Common difference of the A.P. (d) = 5/4 - 3/4
= (5-3)/4
= 2/4
Now, as we know,
a_n = a + (n - 1)d
So, for 9^th term,
a₉ = a + (9 - 1)d
= 3/4 + (8)(2/4)
= 3/4 + 16/4
= 19/4
Therefore, the 9^th term of the given A.P. is a₉ = 19/4.

Ques.2. Find:
(i) Which term of the A.P. 3, 8, 13, ... is 248?
(ii) Which term of the A.P. 84, 80, 76, ... is 248?
(iii) Which term of the A.P. 4, 9, 14, ... is 254?
(iv) Which term of the A.P. 21, 42, 63, 84, ... is 420?
(v) Which term of the A.P. 121, 117, 113, ... is its first negative term?
(vi) Which term of the A.P. –7, –12, –17, –22,... will be –82? Is –100 any term of the A.P?
Ans. In the given problem, we are given an A.P and the value of one of its term.
We need to find which term it is (n)
So here we will find the value of n using the formula, a_n = a + (n - 1)d
(i) Here, A.P is 3, 8, 13, ...
a_n = 248
a = 3
Now,
Common difference (d) = a₁ - a
= 8 - 3
= 5
Thus, using the above mentioned formula 
a_n = a + (n - 1)d
248 = 3 + (n - 1)5
248 - 3 = 5n - 5
245 + 5  = 5n
n = 250/5
n = 50
Thus, n = 50
Therefore 248 is the 50^th term of the given A.P.
(ii) Here, A.P is 84, 80, 76,...
a_n = 0
a = 84
Now,
Common difference (d) = a₁ - a
= 80 - 84
= -4
Thus, using the above mentioned formula
a_n = a + (n - 1)d
0 = 84 + (n - 1)(-4)
0 = 84 - 4n + 4
0 = 88 - 4n
4n = 88
On further simplifying, we get,
n = 88/4
n = 22
Thus, n = 22
Therefore 84 is the 22^nd term of the given A.P.
(iii) Here, A.P is 4, 9, 14, ...
a_n = 254
a = 4
Now,
Common difference (d) = a₁ - a
= 9 - 4
= 5
Thus, using the above mentioned formula
a_n = a + (n - 1)d
254 = 4 + (n - 1)5
254 - 4 = 5n - 5
250 + 5 = 5n
n = 255/5
n = 51
Thus, n = 51
Therefore 254 is the 51^st term of the given A.P.
(iv) Here, A.P is 21, 42, 63, 84, ...
a_n = 420
a = 21
Now,
Common difference (d) = a₁ - a
= 42 - 21
= 21
Thus, using the above mentioned formula
a_n = a + (n - 1)d
420 = 21 + (n - 1)21
420 - 21 = 21n - 21
399 + 21 = 21n
n = 420/21
n = 20
Thus, n = 20
Therefore 420 is the 20^th term of the given A.P .
(v) Here, A.P is 121, 117, 113, ...
We need to find first negative term of the A.P.
a = 121
Now,
Common difference (d) = a₁ - a
= 117 - 121 = -4
Now, we need to find the first negative term,
a_n < 0
121 + (n - 1)(-4) < 0
121 - 4n + 4 < 0
125 - 4n < 0
4n > 125
Further simplifying, we get,
n > 125/4
n > 31(1/4)
n ≥ 32 (as n is a natural number)
Thus, n = 32
Therefore, the first negative term is the 32^nd term of the given A.P.
(vi)  –7, –12, –17, –22,...
Given:
Given:
a = −7
d = −12 − (−7)  
= −12 + 7 = −5
a_n = −82
Now,
a_n = a + (n − 1)d
⇒ −82 = −7 + (n − 1)(−5)
⇒ −82 = −7 − 5n + 5
⇒ −82 = −2 − 5n
⇒ −82 + 2 = −5n
⇒ −80 = −5n
⇒ n = −80/−5
⇒ n = 16
Hence, 16^th term of the A.P. is −82.
If a_n = −100
a_n = a + (n − 1)d
⇒ −100 = −7 + (n − 1)(−5)
⇒ −100 = −7 − 5n + 5
⇒ −100 = −2 − 5n
⇒ −100 + 2 = −5n
⇒ −98 = −5n
⇒ n = -98/-5
⇒ n =19.6 which is not possible since n is a natural number.
Hence, −100 is not the term of this A.P.