question_answer

The 14th term of an AP is twice its 8th term. If its 6th term is\[8\], then find the sum of its first 20 terms.

Answer:

In the given AP, let first term \[=a\] and common difference \[=d\]

Then,                \[{{T}_{n}}=a+(n-1)d\]

\[\Rightarrow {{T}_{14}}=a+(14-1)d=a+13d\]

and                   \[{{T}_{8}}=a+(8-1)d=a+7d\]

Now,                 \[{{T}_{14}}=2{{T}_{8}}\]                                          (Given)

\[a+13d=2(a+7d)\]

\[a+13d=2a+14d\]

\[a=-d\]                                                 ?(i)

Also,                 \[{{T}_{6}}=a+(6-1)d\]

\[\Rightarrow a+5d=-8\]                                      ?(ii)

Putting the value of a from eq. (i), we get

\[-d+5d=-8\]

\[4d=-8\]

\[d=-2\]

Substituting \[d=-2\] in eq. (ii), we get

\[a+5(-2)=-8\]

\[a=10-8\]

\[a=2\]

\[\therefore \]  Sum of first 20 terms is

\[{{S}_{20}}=\frac{n}{2}[2a+(n-1)d]\]

\[=\frac{20}{2}[2\times 2+(20-1)(-2)]\]

\[=10[4-38]\]

\[=-340\]