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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 $-340$.
Updated on: 10-Oct-2022
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