Exercise 5.2
Question 13:
How many three digit numbers are divisible by 7?
First three-digit number that is divisible by 7 = 105
Next number = 105 + 7 = 112
Therefore, A.P will be 105, 112, 119,…
The last three digit number = 999
Therefore last three digit number divisible by 7 = 999 ─ 5 = 994.
When we divide it by 7, the remainder will be 5.
Clearly, 999 − 5 = 994 is last three-digit number that is divisible by 7.
The series is as follows.
105, 112, 119, …, 994
Let 994 be the nth term of this A.P.
a = 105
d = 7
an = 994
n = ?
an = a + (n − 1) d
994 = 105 + (n − 1) 7
889 = (n − 1) 7
(n − 1) = 127
n = 128
Therefore, 128 three-digit numbers are divisible by 7.
Question 14:
How many multiples of 4 lie between 10 and 250?
Therefore, 12, 16, 20, 24, … will be series
All these are divisible by 4 and thus, all these are terms of an A.P.
Therefore, 250 − 2 = 248 is
divisible by 4.
Let 248 be the nth term of this A.P.
Therefore, there are 60 multiples of 4 between 10 and 250.
Question 15:
For what value of n, are the nth terms of two APs 63, 65, 67, and 3, 10, 17, … equal.
First A.P 63, 65, 67, …
a = 63
d = a2 − a1 = 65 − 63 = 2
nth term of this A.P. = an = a + (n − 1) d
an= 63 + (n − 1) 2 = 63 + 2n − 2
an = 61 + 2n ……………….(1)
Second A.P 3, 10, 17, …
a = 3
d = a2 − a1 = 10 − 3 = 7
nth term of this A.P. = 3 + (n − 1) 7
an = 3 + 7n − 7
an = 7n − 4 ………………..(2)
It is given that, nth term of these A.P.s are equal to each other.
Equating equations (1) and (2),
61 + 2n = 7n − 4
61 + 4 = 5n
5n = 65
n = 13
Therefore, 13th terms of both these A.P.s are equal to each other.