Find the sum of first $n$ term of an $A . P$ is $c n^{2}$ , which $n t h$ term is $t_{n} = 5 + 6 n n \in N$ .

\frac{n (4 n^{2} - 1) c^{2}}{6}

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3 n^{2} + 8 n

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\frac{n (4 n^{2} - 1) c^{2}}{3}

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\frac{n (4 n^{2} + 1) c^{2}}{6}

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Solution

The correct option is B $3 n^{2} + 8 n$
$\begin{matrix} T h e g i v n e s e q u e n c e o f a n A . P w i t h t_{n} = 5 + 6 n, n \in N ∴ T h e f i r s t t e r m, t_{1} = 5 + 6 (1) = 11 a n d t h e l a s t t e r m i s t h e t h e n t h t e r m t_{n} = 5 + 6 n N o w, S u m o f n t e r m o f A . P i s g i v n b y S_{n} = \frac{n}{2} [t_{1} + t_{n}] = \frac{n}{2} (11 + 5 + 6 n) = n (8 + 3 n) = 3 n^{2} + 8 H e n c e, t h e o p t i o n B i s t h e c o r r e c t a n s w e r . \end{matrix}$

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