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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 =\[\frac{n}{2}\left[ 2 \times 3 + \left( n - 1 \right)3 \right] + \frac{n}{2}\left[ 2 \times 5 + \left( n - 1 \right)4 \right] + \frac{n}{2}\left[ 2 \times 7 + \left( n - 1 \right)5 \right]\]
\[= \frac{n}{2}\left[ \left( 6 + 3n - 3 \right) + \left( 10 + 4n - 4 \right) + \left( 14 + 5n - 5 \right) \right]\]
\[ = \frac{n}{2}\left[ 12n + 18 \right]\]
\[ = 3n\left( 2n + 3 \right)\]
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