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

In the given A.P series let first term be 𝑎 and common difference be d.

$\begin{array}{r}{a}_n=a+(n-1)d\\ a_{14}=a+(14-1)d\\ a_{14}=a+13d\\ \text{ and }{a}_8=a+(8-1)d\\ a_8=a+7d\end{array}$

It is given that 𝑎₁₄ = 2𝑎₈

$\begin{array}{r}a+13d=2(a+7d)\\ a+13d=2a+14d\\ 2a-a=13d-14d\\ a=-d\dots \dots \dots \dots \dots \left(1\right)\end{array}$

It is also given that 𝑎₆ =− 8

$\begin{array}{r}a+(6-1)d=-8\dots \dots \dots ..\text{ (ii) }\\ a+5d=-8\\ -d+5d=-8\left[\text{ putting the value of a from }\right(i\left)\right]\\ 4d=-8\\ d=-2\end{array}$

Substituting 𝑑 =− 2 in equation (ii) 

we get, 𝑎 = 2 Sum of first 20 terms is:

$\begin{array}{r}{S}_n=\frac{n}{2}[2a+(n-1\left]d\right]\\ =\frac{20}{2}\left[2\right(2)+(20-1\left)\right(-2\left)\right]\\ =10[4-38]\\ =-340\end{array}$

Hence, the sum of the first 20 terms is -340