Question

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

Answer 1

Let a be the first term and d be the common difference of the given A.P.
Given :
$a_7=\frac{1}{9}$
$a_9=\frac{1}{7}$
$a_7=a+(7-1)d=\frac{1}{9}$
$\Rightarrow a+6d=\frac{1}{9}$ ....(1)
$a_9=a+(9-1)d=\frac{1}{7}$
$\Rightarrow a+8d=\frac{1}{7}$ ....(2)
Subtracting equation (1) from (2), we get :
$2d=\frac{2}{63}$
$\Rightarrow d=\frac{1}{63}$
Putting $d=\frac{1}{63}$ in equation (1), we get :

a + ($6\times \frac{1}{63}$) = $\frac{1}{9}$

⇒ a = $\frac{1}{63}$

∴ $a_{63}$ =a + (63 - 1) d = $\frac{1}{63}$ +62($\frac{1}{63}$) =$\frac{63}{63}$ = 1

Thus, the 63rd term of the given A.P. is 1.