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Sum-0-7-0-71-0-72-to-100-terms-




Question Number 6328 by Rasheed Soomro last updated on 24/Jun/16
Sum 0.7+0.71+0.72+....to 100 terms.
$$\mathrm{Sum}\:\mathrm{0}.\mathrm{7}+\mathrm{0}.\mathrm{71}+\mathrm{0}.\mathrm{72}+….\mathrm{to}\:\mathrm{100}\:\mathrm{terms}. \\ $$
Commented by FilupSmith last updated on 24/Jun/16
difference d = 0.01  arithmetic series:  ∴ S=(n/2)(2a+(n−1)d)  n=100, a=0.7, d=0.01  S=((100)/2)(1.4+99×0.01)  S=50(1.4+0.99)  S=50(2.39)  S=119.5
$$\mathrm{difference}\:{d}\:=\:\mathrm{0}.\mathrm{01} \\ $$$$\mathrm{arithmetic}\:\mathrm{series}: \\ $$$$\therefore\:{S}=\frac{{n}}{\mathrm{2}}\left(\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right) \\ $$$${n}=\mathrm{100},\:{a}=\mathrm{0}.\mathrm{7},\:{d}=\mathrm{0}.\mathrm{01} \\ $$$${S}=\frac{\mathrm{100}}{\mathrm{2}}\left(\mathrm{1}.\mathrm{4}+\mathrm{99}×\mathrm{0}.\mathrm{01}\right) \\ $$$${S}=\mathrm{50}\left(\mathrm{1}.\mathrm{4}+\mathrm{0}.\mathrm{99}\right) \\ $$$${S}=\mathrm{50}\left(\mathrm{2}.\mathrm{39}\right) \\ $$$${S}=\mathrm{119}.\mathrm{5} \\ $$

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