The 7th term of an A.P. is 32 and its 13th term is 62. Find the A.P.

Given:

The 7th term of an A.P. is 32 and its 13th term is 62.

To do:

We have to find the A.P.

Solution:

Let the common difference of the A.P. be $d$ and the first term be $a$.

We know that,

nth term of an A.P. $a_n=a+(n-1)d$

Therefore,

$a_{7}=a+(7-1)d$

$32=a+6d$

$a=32-6d$......(i)

$a_{13}=a+(13-1)d$

$62=a+12d$

$62=32-6d+12d$         (From (i))

$6d=62-32$

$d=\frac{30}{6}$

$d=5$......(ii)

This implies,

$a_1=a=32-6(5)=32-30=2$

$a_{2}=a+(2-1)d=a+d=2+5=7$

$a_3=a+(3-1)d=a+2d=2+2(5)=2+10=12$

$a_4=a+(4-1)d=a+3d=2+3(5)=2+15=17$

Hence, the required A.P. is $2, 7, 12, 17......$