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  2. How many terms of the series 51+54+57+....are required to get a sum of 810
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How many terms of the series 51+54+57+....are required to get a sum of 810

Answers (1)
S safeer

The given series is an AP

The sum of n terms of an AP is given by 

\\s_n=\frac{n}{2}(2a+(n-1)d)

a is the first term and d is the common difference.

\\810=\frac{n}{2}(102+(n-1)3)\\\\1620=102n+3n^2-3n\\3n^2+99n-1620=0\\n^2+33n-540=0\\\\n=\frac{-33\pm\sqrt{33^2+4\times1\times540}}{2}\\\\n=\frac{-33\pm\sqrt{3249}}{2}=\frac{-33\pm{57}}{2}\\\text{we have to take only the positive value}\\n=12

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