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Question

# How many terms of the A.P. −6, $-\frac{11}{2}$, −5, ... are needed to give the sum −25?

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Solution

## $\mathrm{Given}:\phantom{\rule{0ex}{0ex}}\mathrm{An} \mathrm{A}.\mathrm{P}.\mathrm{with}a=-6\mathrm{and}d=-\frac{11}{2}-\left(-6\right)=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{S}_{n}=-25\phantom{\rule{0ex}{0ex}}\therefore -25=\frac{n}{2}\left[2×\left(-6\right)+\left(n-1\right)\frac{1}{2}\right]\phantom{\rule{0ex}{0ex}}⇒-25=\frac{n}{2}\left[-12+\frac{n}{2}-\frac{1}{2}\right]\phantom{\rule{0ex}{0ex}}⇒-50=n\left[\frac{n}{2}-\frac{25}{2}\right]\phantom{\rule{0ex}{0ex}}⇒-100=n\left(n-25\right)\phantom{\rule{0ex}{0ex}}⇒{n}^{2}-25n+100=0\phantom{\rule{0ex}{0ex}}⇒\left(n-20\right)\left(n-5\right)=0\phantom{\rule{0ex}{0ex}}⇒n=20\mathrm{or}n=5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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