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# 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 $-\frac{4}{15}$.

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