Question

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

Answer 1

$T_n$ = a+ (n-1)d
Now, $T_7$ = a+6d = 32---------------(1)
Also given $T_{13}$ = a+ 12d =62-------------------(2)
Subtracting (1) from (2) we have
$\Rightarrow$ -6d = (-30)
$\Rightarrow$ d = 5

Putting the value of 'd ' in (1)
$\Rightarrow$ a + 6 (5) = 32
$\Rightarrow$ a + 30 = 32

$\Rightarrow$ a = 32- 30
$\Rightarrow$ a = 2, d = 5

So, the required A.P is 2, 7, 12, 17,22.......