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

Nth term = a+ ( n-1 ) d
( where a is the first term, n is the no. of terms
and d is the difference between two consecutive terms . )

seventh term = $a_7$= $\frac{1}{9}$
ninth term = $a_9$ = $\frac{1}{7}$
$a_7$ = a + 6d ( Eq 1 )
$a_9$= a+ 8d ( Eq 2 )
by subtracting ( 1 ) from ( 2 )
$\Rightarrow a+8d-a-6d=\frac{1}{7}-\frac{1}{9}$

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

$\Rightarrow d=\frac{1}{63}$
by putting value of d in Eq. ( 2 )
a + $8\times \frac{1}{63}=\frac{1}{7}$
a + $\frac{8}{63}=\frac{1}{7}$
a = $\frac{1}{7}-\frac{8}{63}$
a = $\frac{1}{63}$
$a_{63}$ = a + 62d
= $\frac{1}{63}+62\times \frac{1}{63}$
= $\frac{1}{63}$ + $\frac{62}{63}$
= 1 ans.