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Question Number 45410 by Rio Michael last updated on 12/Oct/18

$${given}\:{the}\:{AP} \\ $$$${a},{a}\:+{d}\:,{a}+\mathrm{2}{d},{a}+\mathrm{3}{d},… \\ $$$${show}\:{that}\: \\ $$$${S}_{{n}} =\:\frac{{n}}{\mathrm{2}}\left\{\left(\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right\}\right. \\ $$

Commented by maxmathsup by imad last updated on 13/Oct/18

$${we}\:{have}\:{u}_{{n}} ={a}\:+{nd}\:\:\:\:\:{let}\:{S}_{{n}} ={u}_{\mathrm{0}} +{u}_{\mathrm{1}} +….+{u}_{{n}−\mathrm{1}} \:\Rightarrow \\ $$$${S}_{{n}} =\frac{{n}}{\mathrm{2}}\left\{\:{u}_{\mathrm{0}} +{u}_{{n}−\mathrm{1}} \right\}=\frac{{n}}{\mathrm{2}}\left\{{a}\:+{a}\:+\left({n}−\mathrm{1}\right){d}\right\}=\frac{{n}}{\mathrm{2}}\left\{\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right\}. \\ $$$$ \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

$${S}_{{n}} ={a}+\left({a}+{d}\right)+\left({a}+\mathrm{2}{d}\right)+..+\left\{{a}+\left({n}−\mathrm{1}\right){d}\right\} \\ $$$${S}_{{n}} =\left\{{a}+\left({n}−\mathrm{1}\right){d}\right\}+\left\{{a}+\left({n}−\mathrm{2}\right){d}\right\}+..+{a} \\ $$$${add}\:{them} \\ $$$$\mathrm{2}{S}_{{n}} =\left[{a}+{a}+\left({n}−\mathrm{1}\right){d}\right]+\left[{a}+{d}+{a}+\left({n}−\mathrm{2}\right){d}\right]+.. \\ $$$$\mathrm{2}{S}_{{n}} =\left[\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right]+\left[\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right]+..{n}\:{terms} \\ $$$$\mathrm{2}{S}_{{n}} ={n}\left[\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right]\left(.{nterms}\right) \\ $$$$\mathrm{2}{S}_{{n}} ={n}\left[\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right] \\ $$$${S}_{{n}} =\frac{{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right] \\ $$$$ \\ $$

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