Question

# Find the sum of the series: 3 + 5 + 7 + 6 + 9 + 12 + 9 + 13 + 17 + ... to 3n terms.

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Solution

## The given sequence i.e., 3 + 5 + 7 + 6 + 9 + 12 + 9 + 13 + 17 +..... to 3n terms. can be rewritten as 3 + 6 + 9 + .... to n terms + 5 + 9 + 13 + .... to n terms + 7 + 12 + 17 + .... to n terms Clearly, all these sequence forms an A.P. having n terms with first terms 3, 5, 7 and common difference 3, 4, 5 Hence, required sum = $\frac{n}{2}\left[2×3+\left(n-1\right)3\right]+\frac{n}{2}\left[2×5+\left(n-1\right)4\right]+\frac{n}{2}\left[2×7+\left(n-1\right)5\right]$ $=\frac{n}{2}\left[\left(6+3n-3\right)+\left(10+4n-4\right)+\left(14+5n-5\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{n}{2}\left[12n+18\right]\phantom{\rule{0ex}{0ex}}=3n\left(2n+3\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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