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

Ques.21. Write the expression a_n- a_k for the A.P. a, a + d, a + 2d, ...
Hence, find the common difference of the A.P. for which
(i) 11^th term is 5 and 13^th term is 79.
(ii) a₁₀ −a₅ = 200
(iii) 20^th term is 10 more than the 18^th term.
Ans. A.P: a, a+d, a+2d …
Here, we first need to write the expression for a_n - a_k
Now, as we know,
a_n = a + (n - 1)d
So, for n^th term,
a_n = a + (n - 1)d
Similarly, for k^th term
a_k = a + (k - 1)d
So,
a_n - a_k = (a + nd - d) - (a + kd - d)
= a + nd - d - a - kd + d
= nd - kd
= (n - k)d
So, a_n - a_k = (n - k)d
(i) In the given problem, we are given 11^th and 13^th term of an A.P.
We need to find the common difference. Let us take the common difference as d and the first term as a.
Here,
a₁₁ = 5
a₁₃ = 79
Now, we will find a₁₁ and a₁₃ using the formula a_n = a + (n - 1)d
So,
a₁₁ = a + (11 - 1)d
5 = a  + 10d ...(1)
Also,
a₁₃ = a + (13 - 1)d
79 = a + 12d ...(2)
Solving for a and d
On subtracting (1) from (2), we get
79 - 5 = (a + 12d) - (a + 10d)
74 = a + 12d - a - 10d
74 = 2d
d = 74/2
d = 37
Therefore, the common difference for the A.P. is d = 37.
(ii) We are given, a₁₀ - a₅ = 200
Here,
Let us take the first term as a and the common difference as d
Now, as we know,
a_n = a + (n - 1)d
Here, we find a₃₀ and a₂₀.
So, for 10^th term,
a₁₀ = a + (10 - 1)d
= a + (9)d
Also, for 5^th term,
a₅ = a + (5 - 1)d
= a + (4)d
So,
a₁₀ - a₅ = (a + 9d) - (a + 4d)
200 = a + 9d - a - 4d
200 = 5d
d = 200/5
d = 40
Therefore, the common difference for the A.P. is d = 4.
(iii) In the given problem, the 20^th term is 10 more than the 18^th term. So, let us first find the 20^th term and 18^th term of the A.P.
Here
Let us take the first term as a and the common difference as d
Now, as we know,
a_n = a + (n - 1)d
So, for 20^th term (n = 20),
a₂₀ = a + (20 - 1)d
= a + 19d
Also, for 18^th term (n = 18),
a₁₈ = a + (18 - 1)d
= a + 17d
Now, we are given,
a₂₀ = a₁₈ + 10
On substituting the values, we get,
a + 19d = a + 17d + 10
19d - 17d = 10
2d = 10
d = 10/2
d = 5
Therefore, the common difference for the A.P. is d = 5.

Ques.22. Find n if the given value of x is the nth term of the given A.P.
(i) 25, 50, 75, 100, ...; x = 1000
(ii) −1, −3, −5, −7, ...; x = −151
(iii) 5(1/2), 11, 16(1/2), 22, ...; x = 550
(iv) 1, 21/11, 31/11, 41/11,..., x = 171/11
Ans. In the given problem, we need to find the number of terms in an A.P
(i) 25, 50, 75, 100 …
We are given,
a_n = 1000
Let us take the total number of terms as n.
So,
First-term (a) = 25
Last term (a_n) = 1000
Common difference (d) = 50 - 25 = 25
Now, as we know,
a_n = a + (n - 1)d
So, for the last term,
1000 = 25 + (n - 1)25
1000 = 25 + 25n - 25
1000 = 25n
n = 1000/25
n = 40
Therefore, the total number of terms of the given A.P. is n = 40.
(ii) -1, -3, -5, -7 …
We are given,
a_n = -151
Let us take the total number of terms as n.
So,
First-term (a) = −1
Last term (a_n) = −151
Common difference (d) = -3 - (-1)
= -3 + 1= -2
Now, as we know,
a_n = a + (n - 1)d
So, for the last term,
-151 = -1 + (n - 1)(-2)
-151 = -1 - 2n + 2
-151 = 1 - 2n
-2n = -151 - 1
On further simplifying, we get,
-2n = -152
n = -152/-2
n = 76
Therefore, the total number of terms of the given A.P. is n = 76.
(iii)  5(1/2), 11, 16(1/2), 22, ...
We are given,
a_n = 550
Let us take the total number of terms as n.
So,
First term (a) = 5(1/2)
Last term (a_n) = 550
Common difference (d) = (11 - 5(1/2))
= 11 - (11/2)

= 11/2
Now, as we know,
a_n = a + (n - 1)d
So, for the last term,

550 = 11/2(n)
n = 550(2)/11
On further simplifying, we get,
n = 1100/11
n = 100
Therefore, the total number of terms of the given A.P. is n = 100.
_{(iv) 1, 21/11, 31/11, 41/11,...} 
We are given,
a_n = 171/11
Let us take the total number of terms as n.
So,
First-term (a) = 1
Last term (a_n) = 171/11
Common difference (d) = (21/11) - 1

Now, as we know,
a_n = a + (n - 1)d
So, for the last term,

On further simplifying, we get,

Therefore, the total number of terms of the given A.P. is n = 17.

Ques.23. The eighth term of an A.P. is half of its second term and the eleventh term exceeds one-third of its fourth term by 1. Find the 15^th term.
Ans. a₈ = (1/2)a₂ and a₁₁ = (1/3)a₄ + 1
⇒ a + 7d =
⇒ a + 13d = 0 ...(i)
And,
a₁₁ = (1/3)a₄ + 1
⇒ a + 10d =  + 1
⇒ 2a + 27d = 1 ...(ii)
Solve (i) and (ii), we get d = 1.
Therefore, a = −13
a₁₅ = a + 14d = −13 + 14 = 1

Ques.24. Find the arithmetic progression whose third term is 16 and seventh term exceeds its fifth term by 12.
Ans. Here, let us take the first term of the A.P as a and the common difference of the A.P as d
Now, as we know,
a_n = a + (n - 1)d
So, for 3^rd term (n = 3),
a₃ = a + (3 - 1)d
16 = a + 2d
a = 16 - 2d ...(1)
Also, for 5^th term (n = 5),
a₅ = a + (5 - 1)d
= a + 4d
For 7^th term (n = 7),
a₇ = a + (7 - 1)d
= a + 6d
Now, we are given,
a₇ = 12 + a₅
a + 6d = 12 + a + 4d
6d - 4d = 12
2d = 12
d = 6
Substituting the value of d in (1), we get,
a = 16 - 2(6)
= 16 - 12 = 4
So, the first term is 4 and the common difference is 6.
Therefore, the A.P. is 4, 10, 16, 22, ...

Ques.25. The 7^th term of an A.P. is 32 and its 13^th term is 62.  Find the A.P.
Ans. Here, let us take the first term of the A.P. as a and the common difference of the A.P as d
Now, as we know,
a_n = a + (n - 1)d
So, for 7^th term (n = 7),
a₇ = a + (7 - 1)d
32 = a + 6d ...(1)
Also, for 13^th term (n = 13),
a₁₃ = a + (13 - 1)d
62 = a + 12d ...(2)
Now, on subtracting (2) from (1), we get,
62 - 32 = (a + 12d)-(a + 6d)
30 = a + 12d - a - 6d
30 = 6d
d = 30/6
d = 5
Substituting the value of d in (1), we get,
32 = a + 6(5)
32 = a + 30
a = 32 - 30
a = 2
So, the first term is 2 and the common difference is 5.
Therefore, the A.P. is 2, 7, 12, 27, ...

Ques.26. Which term of the A.P. 3, 10, 17, ... will be 84 more than its 13th term?Ans. In the given problem, let us first find the 13^th term of the given A.P.
A.P. is 3, 10, 17 …
Here,
First term (a) = 3
Common difference of the A.P. (d) = 10 - 3 =7
Now, as we know,
a_n = a + (n - 1)d
So, for 13^th term (n = 13),
a₁₃ = 3 + (13 - 1)(7)
= 3 + 12(7)
= 3 + 84
= 87
Let us take the term which is 84 more than the 13^th term as a_n. So,
a_n = 84 + a₁₃
= 84 + 87
= 171
Also, a_n = a + (n - 1)d
171 = 3 + (n - 1)7
171 = 3 + 7n - 7
171 = -4 + 7n
171 + 4 = 7n
Further simplifying, we get,
175 = 7n
n = 175/7
n = 25
Therefore, the 25^th term of the given A.P. is 84 more than the 13^th term.

Ques.27. Two arithmetic progression have the same common difference. The difference between their 100th terms is 100, What is the difference between their 1000th terms?
Ans. Here, we are given two A.P sequences which have the same common difference. Let us take the first term of one A.P. as a and of other A.P. as a’
Also, it is given that the difference between their 100^th terms is 100.
We need to find the difference between their 100^th terms
So, let us first find the 100^th terms for both of them.
Now, as we know,
a_n = a + (n - 1)d
So, for 100^th term of first A.P. (n = 100),
a₁₀₀ = a + (100 - 1)d
= a + 99d
Now, for 100^th term of second A.P. (n = 100),
a'₁₀₀ = a' + (100 - 1)d
= a' + 99d
Now, we are given,
a₁₀₀ - a'₁₀₀ = 100
On substituting the values, we get,
a + 99d - a' - 99d = 100
a - a' = 100 ...(1)
Now, we need the difference between the 1000^th terms of both the A.P.s
So, for 1000^th term of first A.P. (n = 1000),
a₁₀₀₀ = a + (1000 - 1)d
= a + 999d
Now, for 1000^th term of second A.P. (n = 1000),
a'₁₀₀₀ = a' + (1000 - 1)d
= a' + 999d
So,
a₁₀₀₀ - a'₁₀₀₀ = (a + 999d) - (a' + 999d)
= a + 999d - a' - 999d
= a - a'
= 100 (Using 1)
Therefore, the difference between the 1000^th terms of both the arithmetic progressions will be 100.

Ques.28. For what value of n, the nth terms of the arithmetic progressions 63, 65, 67, ... and 3, 10, 17, ... are equal?
Ans. Here, we are given two A.P. sequences. We need to find the value of n for which the n^th terms of both the sequences 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 63, 65, 67 …
Here,
First-term (a) = 63
Common difference of the A.P. (d) = 65 - 63 = 2
Now, as we know,
a_n = a + (n - 1)d
So, for n^th term,
a_n = 63 + (n - 1)2
= 63 + 2n - 2
= 61 + 2n ...(1)
Second A.P. is 3, 10, 17 …
Here,
First-term (a) = 3
Common difference of the A.P. (d) = 10 - 3 = 7
Now, as we know,
a_n = a + (n - 1)d
So, for n^th term,
a_n = 3 + (n - 1)7
= 3 + 7n - 7
= -4 + 7n ...(2)
Now, we are given that the n^th terms for both the A.P. sequences are equal, we equate (1) and (2),
61 + 2n = -4 + 7n
2n - 7n = -4 - 61
-5n = -65
n = -65/-5
n = 13
Therefore, n = 13.

Ques.29. How many multiples of 4 lie between 10 and 250?
Ans. In this problem, we need to find out how many multiples of 4 lie between 10 and 250.
So, we know that the first multiple of 4 after 10 is 12 and the last multiple of 4 before 250 is 248. Also, all the terms which are divisible by 4 will form an A.P. with the common difference of 4.
So here,
First-term (a) = 12
Last term (a_n) = 248
Common difference (d) = 4
So, let us take the number of terms as n
Now, as we know,
a_n = a + (n - 1)d
So, for the last term,
248 = 12 + (n - 1)4
248 = 12 + 4n - 4
248 = 8 + 4n
248 - 8 = 41
Further simplifying,
240 = 4n
n = 240/4
n = 60
Therefore, the number of multiples of 4 that lie between 10 and 250 is 60.

Ques.30. How many three-digit numbers are divisible by 7?
Ans. In this problem, we need to find out how many numbers of three digits are divisible by 7.
So, we know that the first three-digit number that is divisible by 7 is 105 and the last three-digit number divisible by 7 is 994. Also, all the terms which are divisible by 7 will form an A.P. with the common difference of 7.
So here,
First-term (a) = 105
Last term (a_n) = 994
Common difference (d) = 7
So, let us take the number of terms as n
Now, as we know,
a_n = a + (n - 1)d
So, for the last term,
994 = 105 + (n - 1)7
994 = 105 + 7n - 7
994 = 98 + 7n
994 - 98 = 7n
Further simplifying,
896 = 7n
n = 896/7
n = 128
Therefore, the number of three-digit terms divisible by 7 is128.