The sum of the first n …

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Sia ? 6 years, 4 months ago
According to the question,we are given that,
{tex}S _ { n } = \frac { 3 n ^ { 2 } } { 2 } + \frac { 5 n } { 2 } = \frac { 3 n ^ { 2 } + 5 n } { 2 }{/tex}
{tex}\Rightarrow S _ { n - 1 } = \frac { 3 ( n - 1 ) ^ { 2 } + 5 ( n - 1 ) } { 2 }{/tex}
{tex}= \frac { 3 \left( n ^ { 2 } - 2 n + 1 \right) + 5 n - 5 } { 2 }{/tex}
{tex}= \frac { 3 n ^ { 2 } - 6 n + 3 + 5 n - 5 } { 2 }{/tex}
{tex}= \frac { 3 n ^ { 2 } - n - 2 } { 2 }{/tex}
Now,
nth term = Tn=Sn-Sn-1={tex}= \frac { 3 n ^ { 2 } + 5 n } { 2 } - \frac { 3 n ^ { 2 } - n - 2 } { 2 } = \frac { 3 n ^ { 2 } + 5 n - 3 n ^ { 2 } + n + 2 } { 2 }{/tex}={tex}\frac{{6n + 2}}{2}{/tex}=3n+1
25th term=T25=3(25)+1=75+1=76.
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