The sum of the series $1 + 6 + 11 + . . . . . n t e r m s$ is

\frac{n}{2} [5 n - 3]

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\frac{n}{3} [5 n - 1]

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\frac{n}{3} [5 n - 4]

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\frac{n}{3} [5 n - 2]

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Solution

The correct option is A $\frac{n}{2} [5 n - 3]$
The series is in AP with $a = 1, d = 5$
The sum upto $n$ terms is given as $\frac{n}{2} [2 (1) + (n - 1) 5] \frac{n}{2} [5 n - 3]$

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$\frac{5^{n + 3} - 6 \times 5^{n + 1}}{9 \times 5^{n} - 5^{n} \times 2^{2}}$ = 19

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