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A number consists of three digits which are in GP the sum of right hand and left hand digits exceeds twice the middle digit by 1 and the sum of left hand and middle digits is two third of the sum of the middle and right hand digits . find the number.

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Solution: Let three digits be $a , ar$ and $ar^2$ , then the number is

$100a+10ar+ar^2$ $............(1)$

Given $a+ar^2=2ar+1$

or $a(r^2-2r+1)=1Rightarrow a(r-1)^2=1$ $..........(2)$

Also given $a+ar=frac23(ar+ar^2)$

$Rightarrow$ $3+3r=2r+2r^2Rightarrow 2r^2-r-3=0$

or $(r+1)(2r-3)=0Rightarrow r=-1,frac32.$

For $r=-1,$ $a=frac1(r-1)^2=frac14 otin I herefore r eq -1$

For $r=frac32,$ $a=frac1(r-1)^2=4$

From Equation $(1)$ , number is $400+10cdot4cdot frac32+4cdot frac94=469.$

Posted by

Deependra Verma

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