Arithmetic progression

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Question

If the sum of the series $2, 5, 8, 11, . . .$ is $60100$ , then $n$ , the number of terms, is

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Solution

We have,
$\begin{array}{l} \frac{n}{2} [2 (2) + (n - 1) 3] = 60100 \\ n (3 n + 1) = 120200 \\ 3 n^{2} + n - 120200 = 0 \\ 3 n^{2} - 600 n + 601 n + 120200 = 0 \\ (n - 200) (3 n + 601) = 0 \Rightarrow n = 200 \end{array}$

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