question_answer

If the sum of the first n terms of an A.P. is \[\frac{1}{2}(3{{n}^{2}}+7n)\], then find its nth term. Hence write its 20th term.

Answer:

Given, \[{{S}_{n}}=\frac{1}{2}(3{{n}^{2}}+7n)\]

Now,                      \[{{S}_{1}}=\frac{1}{2}[3{{(1)}^{2}}+7(1)]=5=a\]           (First term)

And,                      \[{{S}_{2}}=\frac{1}{2}[3{{(2)}^{2}}+7(2)]=13\]

Second term        \[({{a}_{2}})=13-5=8\]

\[\therefore a=5,d=3\]

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

\[=5+(n-1)(3)\]

\[=5+3n-3\]

\[\therefore {{T}_{n}}=3n+2\]

And                  \[{{T}_{20}}=5+(20-1)3\]

\[=5+19\times 3\]

\[\therefore {{T}_{20}}=62\]