Question

If the seventh term of an A.P is $1/9$ and its ninth term is $1/7$, find its $63$rd term.

Answer 1

Let First term of $A.P=a$

And common diff $=d$

then Given us

$T_{7^{th}}=\frac{1}{9}=a+(7-1)d=a+6d.....\left(1\right)$

and $T_{9^{th}}=\frac{1}{7}=a+(9-1)d=a+8d.......\left(2\right)$

Eqn $\left(2\right)-$ Eq $\left(1\right)$

$\frac{1}{7}-\frac{1}{9}=2d$

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

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

Ans From Eqn $\left(1\right)$

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

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

Now $T_{{63}^{th}}=a+(63-1)d$

$=\frac{1}{63}+62\left(\frac{1}{63}\right)$

$T_{{63}^{th}}=1$