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Question

A number consists of three digits in G.P. The sum of the 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 sum of digits of number?

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Solution

Let the three digits be a,ar,ar2 then the number will be
100a+10ar+ar2(1)
Given, a+ar2=2ar+1
a(r22r+1)=1
a(r1)2=1(2)
Also given, a+ar=23(ar+ar2)
3+3r=2r+2r2
2r2r3=0
(r+1)(2r3)=0
r=1,3/2
for r=1,a=1(r1)2=141 r1
for r=3/2,a=1(321)2=4 {From(2)}
From (1), number is 400+10.4.32+4.94=469

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