If the 5th term of an A.P. is 31 and 25th term is 140 more than the 5th term, find the A.P.

Given:

The 5th term of an A.P. is 31 and 25th term is 140 more than the 5th term.

To do:

We have to find the A.P.

Solution:

Let the required A.P. be $a, a+d, a+2d, ......$

Here,

$a_1=a, a_2=a+d$ and Common difference $=a_2-a_1=a+d-a=d$

We know that,

$a_n=a+(n-1)d$

Therefore,

$a_{5}=a+(5-1)d$

$31=a+4d$

$a=31-4d$.....(i)

$a_{25}=a+(25-1)d$

$=a+24d$

According to the question,

$a_{25}=a_5+140$

$a+24d=31+140$

$31-4d+24d=171$

$20d=171-31$

$20d=140$

$d=\frac{140}{20}$

$d=7$

This implies,

$a=31-4(7)$

$=31-28$

$=3$

Therefore,

$a_1=3$

$a_2=a+d=3+7=10$

$a_3=a+2d=3+2(7)=3+14=17$

$a_4=a+3d=3+3(7)=3+21=24$

Hence, the required A.P. is $3, 10, 17, 24,......$