Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-7) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Ques.41. The 24^th term of an A.P. is twice its 10^th term. Show that its 72^nd term is 4 times its 15^th term.
Ans. Let a be the first term and d be the common difference.
We know that, n^th term = a_n = a + (n − 1)d
According to the question,
a₂₄ = 2a₁₀
⇒ a + (24 − 1)d = 2(a + (10 − 1)d)
⇒ a + 23d = 2a + 18d
⇒ 23d − 18d = 2a − a
⇒ 5d = a
⇒ a = 5d .... (1)
Also,
a₇₂ = a + (72 − 1)d
= 5d + 71d [From (1)]
= 76d ..... (2)
and
a₁₅ = a + (15 − 1)d
= 5d + 14d [From (1)]
= 19d ..... (3)
On comparing (2) and (3), we get
76d = 4 × 19d
⇒ a₇₂ = 4 × a₁₅
Thus, 72^nd term of the given A.P. is 4 times its 15^th term.

Ques.42. Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.
Ans. Since, the number is divisible by both 2 and 5, means it must be divisible by 10.
In the given numbers, first number that is divisible by 10 is 110.
Next number is 110 + 10 = 120.
The last number that is divisible by 10 is 990.
Thus, the progression will be 110, 120, ..., 990.
All the terms are divisible by 10, and thus forms an A.P. having first term as 110 and the common difference as 10.
We know that, n^th term = a_n = a + (n − 1)d
According to the question,
990 = 110 + (n − 1)10
⇒ 990 = 110 + 10n − 10
⇒ 10n = 990 − 100
⇒ 10n = 890
⇒ n = 89
Thus, the number of natural numbers between 101 and 999 which are divisible by both 2 and 5 is 89.

Ques.43. If the seventh term of an A.P. is 1/9 and its ninth term is 1/7, find its (63)^rd term.
Ans. Let a be the first term and d be the common difference.
We know that, n^th term = a_n = a + (n − 1)d
According to the question,
a₇ = 1/9
⇒ a + (7 − 1)d = 1/9
⇒ a + 6d = 1/9 .... (1)
Also, a₉ = 1/7
⇒ a + (9 − 1)d = 1/7
⇒ a + 8d = 1/7 ....(2)
On Subtracting (1) from (2), we get
8d − 6d = 1/7 - 1/9
⇒ 2d =
⇒ 2d = 2/63
⇒ d = 1/63
⇒ a = 1/9 - 6/63 [From (1)]
⇒ a =
⇒ a = 1/63
∴ a₆₃ = a + (63 − 1)d
= 1/63 + 62/63
= 63/63
= 1
Thus, (63)^rd term of the given A.P. is 1.

Ques.44. The sum of 5^th and 9^th terms of an A.P. is 30. If its 25^th term is three times its 8^th term, find the A.P.
Ans. Let a be the first term and d be the common difference.
We know that, n^th term = a_n = a + (n − 1)d
According to the question,
a₅ + a₉ = 30
⇒ a + (5 − 1)d + a + (9 − 1)d = 30
⇒ a + 4d + a + 8d = 30
⇒ 2a + 12d = 30
⇒ a + 6d = 15 .... (1)
Also, a₂₅ = 3(a₈)
⇒ a + (25 − 1)d = 3[a + (8 − 1)d]
⇒ a + 24d = 3a + 21d
⇒ 3a − a = 24d − 21d
⇒ 2a = 3d
⇒ a = (3/2)d ....(2)
Substituting the value of (2) in (1), we get
(3/2)d + 6d = 15
⇒ 3d + 12d = 15 × 2
⇒ 15d = 30
⇒ d = 2
⇒ a = (3/2) x 2 [From (1)]
⇒ a = 3
Thus, the A.P. is 3, 5, 7, 9, .... .

Ques.45. Find where 0 (zero) is a term of the A.P. 40, 37, 34, 31, ..... .
Ans. Let a be the first term and d be the common difference.
We know that, n^th term = a_n = a + (n − 1)d
It is given that a = 40, d = −3 and a_n = 0
According to the question,
⇒ 0 = 40 + (n − 1)(−3)
⇒ 0 = 40 − 3n + 3
⇒ 3n = 43
⇒ n = 43/3 .... (1)
Here, n is the number of terms, so must be an integer.
Thus, there is no term where 0 (zero) is a term of the A.P. 40, 37, 34, 31, ..... .

Ques.46. Find the middle term of the A.P. 213, 205, 197, ...., 37.
Ans. Let a be the first term and d be the common difference.
We know that, n^th term = a_n = a + (n − 1)d
It is given that a = 213, d = −8 and a_n = 37
According to the question,
⇒ 37 = 213 + (n − 1)(−8)
⇒ 37 = 213 − 8n + 8
⇒ 8n = 221 − 37
⇒ 8n = 184
⇒ n = 23 .... (1)
Therefore, total number of terms is 23.
Since, there are odd number of terms.
So, Middle term will be  term, i.e., the 12^th term.
∴ a₁₂ = 213 + (12 − 1)(−8)
= 213 − 88
= 125
Thus, the middle term of the A.P. 213, 205, 197, ...., 37 is 125.

Ques.47. If the 5^th term of an A.P. is 31 and 25^th term is 140 more than the 5^th term, find the A.P.
Ans. Let a be the first term and d be the common difference.
We know that, n^th term = a_n = a + (n − 1)d
According to the question,
a₅ = 31
⇒ a + (5 − 1)d =  31
⇒ a + 4d =  31
⇒ a = 31 − 4d .... (1)
Also, a₂₅ = 140 + a₅
⇒ a + (25 − 1)d = 140 + 31
⇒ a + 24d = 171 .... (2)
On substituting the values of (1) in (2), we get
31 − 4d + 24d = 171
⇒ 20d = 171 − 31
⇒ 20d = 140
⇒ d = 7
⇒ a = 31 − 4 × 7 [From (1)]
⇒ a = 3
Thus, the A.P. is 3, 10, 17, 24, .... .

Ques.48. Find the sum of two middle terms of the A.P. : -(4/3), -1, (-2)/3, (-1)/3, ....., 4(1/3).
Ans. a = -(4/3), d = 1/3, a_n = 13/3
⇒ a + (n − 1)d = 13/3
⇒ (-(4/3)) + (n − 1)(1/3) = 13/3
⇒ 13 = −4 + n − 1
⇒ n = 18
Midlle terms are (n/2)^th and  (n/2) +(1)^th , i.e 9th and 10th terms.
a₉ = a + 8d =
a₁₀ = a + 9d =
∴ a₉ + a₁₀ = 9/3 = 3

Ques.49. If (m + 1)^th term of an A.P is twice the (n + 1)^th term, prove that (3m + 1)^th term is twice the (m + n + 1)^th term.
Ans. Here, we are given that (m+1)^th term is twice the (n+1)^th term, for a certain A.P. Here, let us take the first term of the A.P. as a and the common difference as d
We need to prove that a 3_{m + 1} = 2a_m+n+1 
So, let us first find the two terms.
As we know,
a_n' = a + (n' - 1)d
For (m+1)^th term (n’ = m+1)
a_m+1 = a + (m + 1 - 1)d
= a + md
For (n+1)^th term (n’ = n+1),
a_n+1 = a + (n + 1 - 1)d
= a + nd
Now, we are given that a_m+1 = 2a_n+1
So, we get,
a + md = 2(a + nd)
a + md = 2a + 2nd
md - 2nd = 2a - a
(m - 2n)d = a ...(1)
Further, we need to prove that the (3m+1)^th term is twice of (m+n+1)^th term. So let us now find these two terms,
For (m+n+1)^th term (n’ = m+n+1),
a_m+n+1 = a + (m + n + 1 - 1 )d
= (m - 2n)d + (m + n)d (Using 1)
= md - 2nd + md + nd
= 2md - nd
For (3m+1)^th term (n’ = 3m+1),
a_{3m + 1} = a + (3m + 1 - 1)d
= (m - 2n)d + 3md (Using 1)
= md - 2nd + 3md
= 4md - 2nd
= 2(2md - nd)
Therefore, a_3m+1 = 2a_m+n+1 
Hence proved.

Ques.50. If an A.P. consists of n terms with first term a and n^th term l show that the sum of the m^th term from the beginning and the m^th term from the end is (a + l).
Ans. In the given problem, we have an A.P. which consists of n terms.
Here,
The first term (a) = a
The last term (a_n) = l
Now, as we know,
a_n = a + (n - 1)d
So, for the m^th term from the beginning, we take (n = m),
a_m = a + (m - 1)d
= a + md - d ...(1)
Similarly, for the m^th term from the end, we can take l as the first term.
So, we get,
a_m' = l - (m - 1)d
= l - md + d ...(2)
Now, we need to prove a_m + a_m' = a + l
So, adding (1) and (2), we get,
a_m + a_m' = (a + md - d)+(l - md + d)
= a + md - d + l - md + d
= a + l
Therefore, a_m + a_m' = a + l
Hence proved.

Ques.51. How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3?
Ans. The given A.P is 11, 15, 19, .....,299.
a = 11
d = 4
a_n = 299
⇒ a + (n - 1)d = 299
⇒ 11 + (n -1) 4 = 299
⇒ n - 14 = 288
⇒ n = 73

Ques.52. Find the 12^th term from the end of the A.P. −2, −4, −6, ........−100.
Ans. Consider the A.P −2, −4, −6, ...., −100.
a = −2, d = −2, a_n = −100
⇒ a + (n − 1)d = −100
⇒ (−2) + (n − 1)(−2) = −100
⇒ 2n = 100
⇒ n = 50
The 12th term from the end will be the 39th term from the starting.
∴ a₃₉ = a + 38d = −2 + 38(−2) = −78

Ques.53. For the A.P.: −3, −7, −11, ....., can we find a₃₀  − a₂₀ withoput actually finding a₃₀ and  a₂₀? Give reasons for your answer.
Ans. Consider the A.P −3, −7, −11, ....
a = −3, d = −4,
a₃₀ − a₂₀ = [a + (30 − 1)d] − [a + (20 − 1)d]
a₃₀ − a₂₀ = a + 29d − a − 19d
∴ a₃₀ − a₂₀ = 10d
= 10(−4)
= −40

Ques.54. Two A.P.s have the same common difference. The first term of one A.P. is 2 and that of the other is 7. The difference between their 10^th terms is the same as the difference between their 21^st terms, which is the same as the difference between any two corresponding terms. Why?
Ans. First term of 1st A.P is 2.
First term of 2nd A.P is 7.
Consider the difference of their 10th terms,
a₁₀ − a'₁₀ = a + 9d − a' − 9d'
= a − a' + 9d − 9d'
= 2 − 7 + 0 [d = d']
= −5
a₂₁ − a'₂₁ = a + 20d − a' − 20d'
= a − a' + 20d − 20d'
= 2 − 7 + 0 [d = d']
= −5
Therefore, a₁₀ − a'₁₀ = a₂₁ − a'₂₁.
The difference between any two corresponding terms of A.P's is same as the difference between their terms.