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# The $14^{th}$ term of an A.P. is twice its $8^{th}$ term. If its $6^{th}$ term is $-8$, then find the sum of its first $20$ terms.

**Given:**The 14th term of an A.P. is twice its 8th term and its 6th term is -8.

**To do:**To find the sum of its first 20 terms.

**Solution:**

Here it is given that,

$a_{14} =2.a_{8}$

And $a_{6} =-8$

Let us assume the first term of the given A.P. is $a$ and its common difference is $d$.

Then its $n^{th}$ term ,

$a_{n} =a+\left( n-1\right) d$

$a_{14} =a+\left( 14-1\right) d$

$a_{14} =a+13d$

Similarly,

$a_{8} =a+7d$

And $a_{6} =a+5d$

$\Rightarrow a+5d=-8\ \ \ ............\left( 1\right) \ \ \left( \because a_{6} =-8\ given\ \right)$

And $a_{14} =2.a_{8}$

$a+13d=2\left( a+7d\right)$

$\Rightarrow a+13d=2a+14d$

$\Rightarrow a+d=0\ \ \ \ .......................\left( 2\right)$

On subtracting $\left( 2\right)$ from $\left( 1\right)$ ,

$a+5d-a-d=-8$

$\Rightarrow 4d=-8$

$\Rightarrow d=-\frac{8}{4} =-2$

On subtituting this value in $\left( 2\right)$ ,

$a=2$

Then sum of n terms, $S_{n} =\frac{n}{2}\left[ 2a+\left( n-1\right) d\right]$

$\therefore$ The sum of its first 20 terms,

$S_{20} =\frac{20}{2}\left[ 2\times 2-2\left( 20-1\right)\right]$

$\Rightarrow S_{20} =10\left( 4-38\right)$

$\Rightarrow S_{20} =-340$

Therefore, The sum of its first $20^{th}$ terms is $-340$.

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