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Question

# If Sn = n(4n + 1) is the sum of n terms of an A.P., then its common difference is __________.

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Solution

## Given: ${S}_{n}=n\left(4n+1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{Sum}\mathrm{of}\mathrm{one}\mathrm{term}={S}_{1}=1\left(4+1\right)=5\phantom{\rule{0ex}{0ex}}⇒{a}_{1}=a=5...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Sum}\mathrm{of}\mathrm{two}\mathrm{term}s={S}_{2}=2\left(8+1\right)=18\phantom{\rule{0ex}{0ex}}⇒{a}_{1}+{a}_{2}=18...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Subtracting}\left(1\right)\mathrm{from}\left(2\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}{a}_{2}=13\phantom{\rule{0ex}{0ex}}⇒a+d=13\phantom{\rule{0ex}{0ex}}⇒d=8\left(\because a=5\right)$ Hence, if Sn = n(4n + 1) is the sum of n terms of an A.P., then its common difference is 8.

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