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

Given:

The seventh term of an AP is $\frac{1}{9}$ and its ninth term is $\frac{1}{7}$.

To do:

We have to find the 63$^{rd}$ term of the given A.P.

Solution:

Let $a$ be the first term and $d$ be the common difference of the given A.P.

We know that,

$n^{th}$  term of an A.P. $a_{n} =a+( n-1) d$

7th term of the given A.P. $a_{7} =a+( 7-1) d$

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

Similarly,

9th term of the given A.P. $a_{9} =a+( 9-1) d$

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

On subtracting $( 2)$  from $( 1)$

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

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

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

On subtituting the value of $d$ in $( 1)$, we get,

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

$\Rightarrow a=\frac{1}{9} -\frac{6}{63}$

$\Rightarrow a=\frac{7-6}{63} =\frac{1}{63}$

63rd term of the given A.P. $a_{63}=a+( 63-1) d$

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

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

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

Hence, the $63^{rd}$ term of the given A.P. is $1$. 

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Updated on: 10-Oct-2022

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