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Question Number 43628 by peter frank last updated on 12/Sep/18

Three numbers form an increasing GP.  If the middle number is doubled, then the  new numbers are in AP. The common ratio  of the GP is

$$\mathrm{Three}\:\mathrm{numbers}\:\mathrm{form}\:\mathrm{an}\:\mathrm{increasing}\:\mathrm{GP}. \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{middle}\:\mathrm{number}\:\mathrm{is}\:\mathrm{doubled},\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{new}\:\mathrm{numbers}\:\mathrm{are}\:\mathrm{in}\:\mathrm{AP}.\:\mathrm{The}\:\mathrm{common}\:\mathrm{ratio} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{GP}\:\mathrm{is} \\ $$

Answered by $@ty@m last updated on 13/Sep/18

ATQ  (a/r),2a,ar are in AP  ⇒4a=(a/r)+ar  ⇒4=(1/r)+r  ⇒4=((1+r^2 )/r)  ⇒r=2±(√3)  Ans.

$${ATQ} \\ $$$$\frac{{a}}{{r}},\mathrm{2}{a},{ar}\:{are}\:{in}\:{AP} \\ $$$$\Rightarrow\mathrm{4}{a}=\frac{{a}}{{r}}+{ar} \\ $$$$\Rightarrow\mathrm{4}=\frac{\mathrm{1}}{{r}}+{r} \\ $$$$\Rightarrow\mathrm{4}=\frac{\mathrm{1}+{r}^{\mathrm{2}} }{{r}} \\ $$$$\Rightarrow{r}=\mathrm{2}\pm\sqrt{\mathrm{3}}\:\:{Ans}. \\ $$

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