question_answer

If \[{{S}_{n}}\], denotes the sum of first n terms of an A.P, prow that \[{{S}_{12}}=3({{S}_{8}}-{{S}_{4}})\]

Answer:

Let a be the first term and d be the common difference

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

Then,                \[{{S}_{12}}=\frac{12}{2}[2a+(12-1)]\]

\[=6(2a+11d)=12a+66d\]

\[{{S}_{8}}=\frac{8}{2}[2a+(8-1)d]\]

\[=4(2a+7d)=8a+28d\]

and,              \[{{S}_{4}}=\frac{4}{2}[2a+(4-1)d]\]

\[=2(2a+3d)=4a+6d\]

Now,

\[3({{S}_{8}}-{{S}_{4}})=3(8a+28d-4a-6d)\]

\[=3(4a+22d)\]

\[=12a+66d\]

\[={{S}_{12}}\]                                    Hence Proved.