Question

If the seventh term of an A.P. is $\frac{1}{9}$ and its ninth term is $\frac{1}{7}$, find its ${63}^{rd}$ term.

Answer 1

We know,

$n^{th}$ term $a_n=a+(n-1)d$

$a=$ First term

$d=$ Common difference

$n=$ number of terms

$a_n=n^{th}$ term

Given

$\frac{1}{9}=a+6d$ ............ (1) 7th term

$\frac{1}{7}=a+8d$ ........... (2) 9th term

Subtracting 1 from 2, we get

$2d=\frac{2}{63}$

$d=\frac{1}{63},a=\frac{1}{63}$

$63rd$ term = $a+62d$

$=\frac{1}{63}+\frac{62}{63}$

$=1$