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If the seventh term of an AP is $\frac {1}{9}$ and its ninth term is $\frac {1}{7}$, find its $63^{rd}$ term.
Given: The seventh term of an AP is $\frac{1}{9}$ and its ninth term is $\frac{1}{7}$.
To do: To find 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.
As known $n^{th}$ term of an A.P., $a_{n} =a+( n-1) d$
Then seventh term of the given A.P. $a_{7} =a+( 7-1) d$
$\Rightarrow \frac{1}{9} =a+6d\ \ \ \ ...........( 1)$
Similarly ninth 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 this value in $( 1)$ ,
$\frac{1}{9} =a+6\times \frac{1}{63}$
$\Rightarrow a=\frac{1}{9} -\frac{6}{63}$
$\Rightarrow a=\frac{7-6}{63} =\frac{1}{63}$
$63^{rd} \ term\ of\ the\ given\ A.P.=a+( 63-1) d$
$a_{63} =\frac{1}{63} +62\times \frac{1}{63}$
$a_{63} =\frac{1}{63} +\frac{62}{63}$
$a_{63} =\frac{63}{63} =1$
Thus $63^{rd}$ term of the given A.P. is 1.
Updated on: 10-Oct-2022
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