Selina Concise Class 10 Math Chapter 10 Arithmetic Progression Solutions
Exercise 10B
(Q1) In an A.P, ten times of its tenth term is equal to thirty times of its 30th term. Find its 40th term.
Solution:
Given that, arithmetic progression.
Ten times of its tenth term.
10 t10 and 30 times of it’s 30th term.
30 t30
∴ 10 t10 = 30 t30
We know that the general term of an arithmetic progression.
Tn = a + (n – 1) d
To find 40th term.
First we have to find t10.
∴ t10 = a + (10 – 1) (∵ n = 10)
T10 = a + 9d —— (1)
And t30 = a + (30 – 1) d (∵ n = 30
T30 = a + 29d —– (2)
∴ 10 (a + 9d) = 30 (a + 29d) (∵ from (1) and (2))
10a + 90d = 30a + 870d
10a – 30a = 870d – 90d
-20a = 780d
-20a – 780d = 0
Multiply by (-)
20a + 780d = 0
20 (a + 39d) = 0
a + 39d = 0
a + (40 – 1) = 0 —– (A) (∵ 39 = 40 – 1)
This is nothing but it’s a term t40.
∴ t40 = a + (40 – 1) d
∴ t40 = 0 (∵ from A)
∴ The 40th term is 0.
(Q2) How many two – digit numbers are divisible by 3?
Solution:
We know that two –digit numbers are divisible by 3.
We have to write a sequence the two-digit numbers are divisibly by 3.
12, 15, 18, 21, 24, 27, 30, —– 99
∴ t1 = 12, t2 = 15, t3 = 18, —-
d1= t2 – t1 = 15 – 12 = 3
d2 = t3 – t2 = 18 – 15 = 3
d3 = t4 – t3 = 21 – 18 = 3
d1 = d2 = d3 = d
We know that,
nth term = 99, a = t1 = 12
And d = 3.
We know that the general term of an arithmetic progression is,
tn = a + (n – 1) d
99 = 12 + (n – 1) 3
99 = 12 + 3n – 3
99 = 9 + 3n
99 – 9 = 3n
90 = 3n
90/3 = n
30 = n
∴ The two-digits numbers divisible by 3 are 30.
(Q3) Which term of arithmetic progression 5, 15, 25, —– will be 130 more than it’s term?
Solution:
Given arithmetic progression is 5, 15, 25, —-
a = t1 = 5, t2 = 15, t3 = 25 —-
t2 – t1 = 15 – 5 = 10
t3 – t2 = 25 – 15 = 10 —
∴ a = 5 and d = 10
Also given that,
130 more than it’s 31st term
∴ tn = t31 + 130
A + (n – 1) d = a + (31 – 1) d + 130
5 + (n – 1) 10 = 5 + (31 – 1) 10 + 130
5 + 10n – 10 = 5 + (30) 10 + 130
5 + 10n – 10 = 5 + 300 + 130
– 5 + 10n = 305 + 130
– 5 + 10n = 435
10n = 435 + 5
10n = 440
n = 440/10
n = 44
∴ The 44th term of given arithmetic progression is 130 more than its 31st term.
(Q4) Find the value of P, if X, 2x + p and 3x + 6 are in A.P.
Solution:
Given that x, 2x + p and 3x + 6 are in arithmetic progression.
∴ There common difference is also same. =
t1 = x, t2 = 2x + P and t3 = 3x + 6
d1 = d2
t2 – t1 = t3 – t2
2x + p – x = 3x + 6 – (2x + p)
x + p = 3x + 6 – 2x – P
x + p = x + 6 – P
x – x + p + p – 6 = 0
2p – 6 = 0
2p = 6
P = 6/2
P = 3
∴ The value of p is 3.
(Q6) How many three – digit numbers are divisible by 87?
Solution:
The three – digit numbers are divisible by 87 are,
174, 261, 348, —— 957
t1 = 174, t2 = 261, t3 = 348 —- tn = 957
∴ t2 – t1 = 261 – 174 = 87 = d1
t3 – t2 = 348 – 261 = 87 = d2
∴ d1 = d2 = d
∴ a = t1 = 174 and d = 87
We know that the general term of arithmetic progression is,
tn = a + (n – 1) d
957 = 174 + (n – 1) 87
957 = 174 + 87n – 87
957 = 87 + 87n
957 – 87 = 87n
870 = 87n
870/87 = n
10 = n
∴ The three – digit numbers are divisible by 8 are 10
(Q7) For what value of n, the nth term of arithmetic progression 63, 65, 67, —– and nth term of arithmetic progression 3, 10, 17, are equal to each other?
Solution:
Given the nth term of arithmetic progression is 63, 65, 67, —- and nth term of arithmetic progression 3, 10, 17, are equal to each other,
Arithmetic progression (1) = 63, 65, 67, —-
a = 63, and t2 – t1 = 65 – 63 = 2 = d
∴ We know that the general term is,
tn = a + (n – 1) d
tn = 63 + (n- 1) 2
= 63 + 2n – 2
tn = 61 + 2n —— (i)
Arithmetic progression (2) = 3, 10, 17, —-
a = t1 = 3 an + t2 – t1 = 10 – 3 = 7 = d
We know that the general term of arithmetic progression.
tn = a + (n – 1) d
tn = 3 + (n – 1) 7
= 3 + 7n – 7
tn = – 4 + 7n —- (2)
from (1) and (2)
61 + 2n = -4 + 7n
61 + 4 = 7n – 2n
65 = 5n
65/5 = n
13 = n
∴ The value of n = 13