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 a
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 thirs of the sum of the middle and right hand digits. Find the number.
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Let the three digits be a, ar and `ar^(2)` ,then number is
`100a+10ar+ar^(2) " " "……(i)"`
Given, `a+ar^(2)=2ar+1`
or `a(r^(2)-2r+1)=1`
or `a(r-1)^(2)=1" " "…….(ii)"`
Also, given `a+ar=(2)/(3)(ar+ar^(2)) implies 3+3r=2r+2r^(2)`
or `2r^(2)-r-3=0` or `(r+1)(2r-3)=0`
`:.r=-1,(3)/(2)`
For `r=-1, a=(1)/((r-1))^(2)=(1)/(4)cancel in I`
`:. rne-1`
For `r=(3)/(2),a=(1)/((3)/(2)-1)^(2)=4" " [" from Eq. (ii) " ]`
From eq.(i), number is `400+10*4*(3)/(2)+4*(9)/(2)=469`
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