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Question

A number consists of three digits which are in GP the sum of the right-hand digits exceeds twice the middle digits by 1 and the sum of the 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

Solution.

Let the three digits be a,ar and ar2.

Then, according to the question,
100a+10ar+ar2(i)a+ar2=2ar+1ar2-2r+1=1a(r-1)2=1(ii)a+ar=23ar+ar23+3r=2r+2r22r2-r-3=0(r+1)(2r-3)=0r=-1,32

For r=-1

a=1(-1-1)2=141r-1.

For r=32

a=132-12=141r-1.

From equation (i) the number is

400+10×4×32+4×94=469

Hence, the number is 469.


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