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Question Number 45422 by Tawa1 last updated on 12/Oct/18

Three consecutive terms of a G.P are the 3rd, 5th and 8th term of an A.P.  Find the common ratio.

$$\mathrm{Three}\:\mathrm{consecutive}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{a}\:\mathrm{G}.\mathrm{P}\:\mathrm{are}\:\mathrm{the}\:\mathrm{3rd},\:\mathrm{5th}\:\mathrm{and}\:\mathrm{8th}\:\mathrm{term}\:\mathrm{of}\:\mathrm{an}\:\mathrm{A}.\mathrm{P}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{common}\:\mathrm{ratio}. \\ $$

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

A+(3−1)D=a  A+(5−1)D=ar  A+(8−1)D=ar^2   {A+4D}−{A+2D}=ar−a  2D=ar−a  D=((ar−a)/2)  A+2D=a  A+ar−a=a  A=2a−ar  A+7D=ar^2   2a−ar+7(((ar−a)/2))=ar^2   4a−2ar+7ar−7a=2ar^2   4−2r+7r−7−2r^2 =0  −2r^2 +5r−3=0  2r^2 −5r+3=0  2r^2 −2r−3r+3=0  2r(r−1)−3(r−1)=0  (r−1)(2r−3)=0  r=1   or r=(3/2)

$${A}+\left(\mathrm{3}−\mathrm{1}\right){D}={a} \\ $$$${A}+\left(\mathrm{5}−\mathrm{1}\right){D}={ar} \\ $$$${A}+\left(\mathrm{8}−\mathrm{1}\right){D}={ar}^{\mathrm{2}} \\ $$$$\left\{{A}+\mathrm{4}{D}\right\}−\left\{{A}+\mathrm{2}{D}\right\}={ar}−{a} \\ $$$$\mathrm{2}{D}={ar}−{a} \\ $$$${D}=\frac{{ar}−{a}}{\mathrm{2}} \\ $$$${A}+\mathrm{2}{D}={a} \\ $$$${A}+{ar}−{a}={a} \\ $$$${A}=\mathrm{2}{a}−{ar} \\ $$$${A}+\mathrm{7}{D}={ar}^{\mathrm{2}} \\ $$$$\mathrm{2}{a}−{ar}+\mathrm{7}\left(\frac{{ar}−{a}}{\mathrm{2}}\right)={ar}^{\mathrm{2}} \\ $$$$\mathrm{4}{a}−\mathrm{2}{ar}+\mathrm{7}{ar}−\mathrm{7}{a}=\mathrm{2}{ar}^{\mathrm{2}} \\ $$$$\mathrm{4}−\mathrm{2}{r}+\mathrm{7}{r}−\mathrm{7}−\mathrm{2}{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$−\mathrm{2}{r}^{\mathrm{2}} +\mathrm{5}{r}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{2}{r}^{\mathrm{2}} −\mathrm{5}{r}+\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{2}{r}^{\mathrm{2}} −\mathrm{2}{r}−\mathrm{3}{r}+\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{2}{r}\left({r}−\mathrm{1}\right)−\mathrm{3}\left({r}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({r}−\mathrm{1}\right)\left(\mathrm{2}{r}−\mathrm{3}\right)=\mathrm{0} \\ $$$${r}=\mathrm{1}\:\:\:{or}\:{r}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Commented by Tawa1 last updated on 12/Oct/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 13/Oct/18

God bless all

$${God}\:{bless}\:{all} \\ $$

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