question_answer

If \[{{S}_{n}}\] denotes the sum of first n terms of an A.P., prove that \[{{S}_{30}}=3[{{S}_{20}}-{{S}_{10}}]\]

Answer:

Let a be the first term and d be the common difference of the A.P.

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

\[\therefore {{S}_{30}}=\frac{30}{2}[2a+(30-1)d]\]

\[=15(2a+29d)\]

\[=30a+435d)\]

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

\[=10[2a+19d]\]

\[=20a+190d\]

And,          \[{{S}_{10}}=\frac{10}{2}[2a+(10-1)d]\]

\[=5[2a+9d]\]

\[=10a+45d\]

Now,     \[3[{{S}_{20}}-{{S}_{10}}]\]\[=3[20a+190d-10a-45d]\]

\[=3[10a+145d]\]

\[=30a+435d\]

\[={{S}_{30}}\]

\[\therefore {{S}_{30}}=3[{{S}_{20}}-{{S}_{10}}]\]                 Hence Proved.