Question

$t_7=\frac{1}{9}$ and $t_9=\frac{1}{7}$ find 63rd term.

Answer 1

Since series is an ${AP}_1$

Let a be first term

and d be common difference, so

$T_7=a+6d=\frac{1}{9}\to 1$

and $T_9=a+8d=\frac{1}{7}\to 2$

On solving $1$ nd $2$, we get

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

and $a=\frac{1}{63}$

So, $T_{63}=a+62d=\frac{1}{63}+\frac{62}{63}=\frac{63}{63}=1$