Question

If $a_7=\frac{1}{9}{a}_9=\frac{1}{7}$ find ${63}^{th}$ term.

Answer 1

$\Rightarrow$ $a_7=\frac{1}{9}$ and $a_9=\frac{1}{7}$

We know, $a_n=a+(n-1)d$

$\Rightarrow$ So, $a_7=\frac{1}{9}$

$\Rightarrow$ $a+6d=\frac{1}{9}$ ------ ( 1 )

$\Rightarrow$ Now, $a_9=\frac{1}{7}$

$\Rightarrow$ $a+8d=\frac{1}{7}$ ------- ( 2 )

Subtracting ( 1 ) from ( 2 ), we get

$8d-6d=\frac{1}{7}-\frac{1}{9}$

$\Rightarrow$ $2d=\frac{9-7}{63}$

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

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

Now from ( 1 ),

$a=\frac{1}{9}-\frac{6}{63}=\frac{1}{9}-\frac{2}{21}=\frac{7-6}{63}=\frac{1}{63}$

$\Rightarrow$ Now, $a_{63}=a+62d$

$\Rightarrow$ $a_{63}=\frac{1}{63}+62\times \frac{1}{63}=\frac{63}{63}=1$

$\therefore$ ${63}^{th}$ term is $1$