Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-3) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 5.24

Ques.3. Find:
(i) Is 68 a term of the A.P. 7, 10, 13, ...?
(ii) Is 302 a term of the A.P. 3, 8, 13, ...?
(iii) Is − 150 a term of the A.P. 11, 8, 5, 2, ...?
Ans. In the given problem, we are given an A.P and the value of one of its term.
We need to find whether it is a term of the A.P or not.
So here we will use the formula, a_n = a + (n - 1)d
(i) Here, A.P is 7, 10, 13, ...
a_n = 68
a = 7
Now,
Common difference (d) = a₁ - a
= 10 - 7
= 3
Thus, using the above mentioned formula, we get,
68 = 7 + (n - 1)3
68 - 7 = 3n - 3
61 + 3 = 3 n
n = 64/3
Since, the value of n is a fraction.
Thus, 68 is not the term of the given A.P
Therefore the answer is NO.
(ii) Here, A.P is 3, 8, 13,...
a_n = 302
a = 3
Now,
Common difference (d) = a₁ - a
= 8 - 3
= 5
Thus, using the above mentioned formula, we get,
302 = 3 + (n - 1)5
302 - 3 = 5n - 5
299 = 5n
n = 299/5
Since, the value of n is a fraction.
Thus, 302 is not the term of the given A.P
Therefore the answer is NO.
(iii) Here, A.P is 11, 8, 5, 2, ...
a_n = -150
a = 11
Now,
Common difference (d) = a₁ - a
= 8 - 11
= - 3
Thus, using the above mentioned formula
-150 = 11 + (n - 1)(-3)
-150 - 11 = -3n + 3
-161 - 3 = -3n
n = -164/-3
Since, the value of n is a fraction.
Thus, -150 is not the term of the given A.P.
Therefore, the answer is NO.

Ques.4. How many terms are there in the A.P.?
(i) 7, 10, 13, ... 43.
(ii) −1, 5/6,2/3,1/2,...10/3.
(iii) 7, 13, 19, ..., 205.
(iv) 18, 15(1/2), 13, ..., −47.
Ans. In the given problem, we are given an A.P.
We need to find the number of terms present in it
So here we will find the value of n using the formula, a_n = a + (n - 1)d
(i) Here, A.P is 7, 10, 13, ... 43.
The first term (a) = 7
The last term (a_n) = 43
Now,
Common difference (d) = a₁ - a
= 10 - 7
= 3
Thus, using the above mentioned formula, we get,
43 = 7 + (n - 1)3
43 - 7 = (3n - 3)
36 + 3 = 3n
n = 39/3
n = 13
Therefore, the number of terms present in the given A.P is 13.
(ii) Here, A.P is −1, 5/6,2/3,1/2,...10/3.
The first term (a) = -1
The last term (a_n) = 10/3
Now,
Common difference (d) = a₁ - a
= -(5/6) - (-1)
= -(5/6) + 1
=
= 1/6
Thus, using the above mentioned formula, we get,


Further solving for n, we get

n = (27/6)(6)
n = 27
Thus, n = 27
Therefore, the number of terms present in the given A.P is 27.
(iii) Here, A.P is 7, 13, 19, ..., 205.
The first term (a) = 7
The last term (a_n) = 205
Now,
Common difference (d) = a₁ - a
= 13 - 7
= 6
Thus, using the above mentioned formula, we get,
205 = 7 + (n - 1)6
205 - 7 = 6n - 6
198 + 6 = 6n
n = 204/6
n = 34
Thus, n = 34
Therefore, the number of terms present in the given A.P is 34.
(iv) Here, A.P is 18, 15(1/2), 13, ..., −47.
The first term (a) = 18
The last term (a_n) = -47
Now,
Common difference (d) = a₁ - a

= (31/2) - 18

= -(5/2)
Thus, using the above mentioned formula, we get,
-47 = 18 + (n - 1)(-(5/2))

Further, solving for n, we get

n = (-135/-5)
n = 27
Thus, n = 27
Therefore, the number of terms present in the given A.P is 27.

Ques.5. The first term of an A.P. is 5, the common difference is 3 and the last term is 80;
Ans. In the given problem, we are given an A.P whose,
First term (a) = 5
Last term (a_n) = 80
Common difference (d) = 3
We need to find the number of terms present in it (n)
So here we will find the value of n using the formula, a_n = a + (n - 1)d
So, substituting the values in the above-mentioned formula
80 = 5 + (n - 1)3
80 - 5 = 3n - 3
75 + 3 = 3n
n = 78/3
n = 26
Thus, n = 26
Therefore, the number of terms present in the given A.P is 26.

Ques.6. The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.
Ans. In the given problem, we are given 6^th and 17^th term of an A.P.
We need to find the 40^th term
Here,
a₆ = 19
a₁₇ = 41
Now, we will find a₆ and a₁₇ using the formula an = a + (n - 1)d
So,
a₆ = a + (6 - 1)d
19 = a + 5d ...(1)
Also,
a₁₇ = a + (17 - 1)d
41 = a + 16d ...(2)
So, to solve for a and d
On subtracting (1) from (2), we get
a + 16d - a - 5d = 41 - 19
11d = 22
d = 22/11
d = 2 ...(3)
Substituting (3) in (1), we get
19 = a + 5(2)
19 - 10 = a
a = 9
Thus,
a = 9
d = 2
n = 40
Substituting the above values in the formula a_n = a + (n - 1)d
a₄₀ = 9 + (40 - 1)2
a₄₀ = 9 + 80 - 2
a₄₀ = 87
Therefore, a₄₀ = 87

Ques.7. The 10^th and 18^th terms of an A.P. are 41 and 73 respectively. Find 26^th term.
Ans. In the given problem, we are given 10^th and 18^th term of an A.P.
We need to find the 26^th term
Here,
a₁₀ = 41
a₁₀ = 73
Now, we will find a₁₀ and a₁₈ using the formula a_n = a + (n - 1)d
So,
a₁₀ = a + (10 - 1)d
41 = a + 9d ...(1)
Also,
a₁₈ = a + (18 - 1)d
73 = a + 17d ...(2)
So, to solve for a and d
On subtracting (1) from (2), we get
8d = 32
d = 32/8
d = 4
Substituting d = 4 in (1), we get
41 = a + 9(4)
41 - 36 = a
a = 5
Thus,
a = 5
d = 4
n = 26
Substituting the above values in the formula, a_n = a + (n - 1)d
a₂₆ = 5 + (26 - 1)4
a₂₆ = 5 + 100
a₂₆ = 105
Therefore, a₂₆ = 105

Ques.8. If the nth term of the A.P. 9, 7, 5, ... is same as the nth term of the A.P. 15, 12, 9, ... find n.
Ans. Here, we are given two A.P. sequences whose n^th terms are equal. We need to find n.
So let us first find the n^th term for both the A.P.
First A.P. is 9, 7, 5 …
Here,
First term (a) = 9
Common difference of the A.P. (d) = 7 - 9 = -2
Now, as we know,
a_n = a + (n - 1)d
So, for n^th term,
a_n = 9 + (n - 1)(-2)
= 9 - 2n + 2
= 11 - 2n ...(1)
Second A.P. is 15, 12, 9 …
Here,
First term (a) = 15
Common difference of the A.P. (d) = 12 - 15 = -3
Now, as we know,
a_n = a + (n - 1)d
So, for n^th term,
a_n = 15 + (n - 1)(-3)
= 15 - 3n + 3
= 18 - 3n ...(2)
Now, we are given that the nth terms for both the A.P. sequences are equal, we equate (1) and (2),
11 - 2n = 18 - 3n
3n - 2n = 18 - 11
n = 7
Therefore, n = 7

Ques.9. Find the 12^th term from the end of the following arithmetic progressions:
(i) 3, 5, 7, 9, ... 201
(ii) 3, 8, 13, ..., 253
(iii) 1, 4, 7, 10, ..., 88
Ans. In the given problem, we need to find the 12^th term from the end for the given A.P.
(i) 3, 5, 7, 9 …201
Here, to find the 12^th term from the end let us first find the total number of terms. Let us take the total number of terms as n.
So,
First term (a) = 3
Last term (a_n) = 201
Common difference (d) = 5 - 3 = 2
Now, as we know,
a_n = a + (n - 1)d
So, for the last term,
201 = 3 + (n - 1)2
201 = 3 + 2n - 2
201 = 1 + 2n
201 - 1 = 2n
Further simplifying,
200 = 2n
n = 200/2
n = 100
So, the 12^th term from the end means the 89^th term from the beginning.
So, for the 89^th term (n = 89)
a₈₉ = 3 + (89 - 1)2
= 3 + (88)2
= 3 + 176
= 179
Therefore, the 12^th term from the end of the given A.P. is 179.
(ii) 3, 8, 13 …253
Here, to find the 12^th term from the end let us first find the total number of terms. Let us take the total number of terms as n.
So,
First term (a) = 3
Last term (a_n) = 253
Common difference, d = 8 - 3 = 5
Now, as we know,
a_n = a + (n - 1)d
So, for the last term,
253 = 3 + (n - 1)5
253 = 3 + 5n - 5
253 = -2 + 5n
253 + 2 = 5n
Further simplifying,
255 = 5n
n = 255/5
n = 51
So, the 12^th term from the end means the 40^th term from the beginning.
So, for the 40^th term (n = 40)
a₄₀ = 3 + (40 - 1)5
= 3 + (39)5
= 3 + 195
= 198
Therefore, the 12^th term from the end of the given A.P. is 198.
(iii) 1, 4, 7, 10 …88
Here, to find the 12^th term from the end let us first find the total number of terms. Let us take the total number of terms as n.
So,
First term (a) = 1
Last term (a_n) = 88
Common difference, d = 4 - 1 = 3
Now, as we know,
a_n = a + (n - 1)d
So, for the last term,
88 = 1 + (n - 1)3
88 = 1 + 3n - 3
88 = - 2 + 3n
88 + 2 = 3n
Further simplifying,
90 = 3n
n = 90/3
n = 30
So, the 12^th term from the end means the 19^th term from the beginning.
So, for the 19^th term (n = 19)
a₁₉ = 1 + (19 - 1)3
= 1 + (18)3
= 1 + 54
= 55
Therefore, the 12^th term from the end of the given A.P. is 55.