# If the $10^{th}$ term of an A.P is $\frac{1}{20}$ and its $20^{th}$ term is $\frac{1}{10}$, then find the sum of its first $200$ terms.

Given: If the $10^{th}$ term of an A.P is $\frac{1}{20}$ and its $20^{th}$ term is $\frac{1}{10}$.
To do: To find the sum of its first $200$ terms.
Solution:
As given, $a_{10}=\frac{1}{20}$
$a_{20}=\frac{1}{30}$
To find $a_{200}=?$
As we know that the n^{th} term of an A.P. is given by-
$a_n=a+(n-1)d$
Here $a$ and $d$ are first term and common difference of A.P. respectively.
Therefore, $a+9d=\frac{1}{20}\ \ \ .....( 1)$
$a+19d=\frac{1}{30}$
$\Rightarrow (a+9d)+10d=\frac{1}{30}$
$\Rightarrow \frac{1}{20}+10d=\frac{1}{30}$        $[From ( 1)]$
$\Rightarrow 10d=\frac{1}{30}-\frac{1}{20}$
$\Rightarrow d=-\frac{1}{600}$
Substituting the value of d in eqn$( 1)$, we have
$a+9( -\frac{1}{600})=\frac{1}{20}$
$\Rightarrow a=\frac{1}{20}+\frac{9}{600}=\frac{39}{600}$
​$\therefore a_{200}=a+199d$
$\Rightarrow a_{200}=\frac{39}{600}+199(-\frac{1}{600})$
$\Rightarrow a_{200}=-\frac{160}{600}$
​$=-\frac{4}{15}$
Thus, the sum of first $200$ terms of the given A.P. is $-\frac{4}{15}$.

Updated on: 10-Oct-2022

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