Question

The sum of three consecutive terms in increasing order of a G.P. is 56. Subtracting 1, 7 and 21 respectively from them, an A.P. is formed. Find those numbers.

Answer 1

Let the GP be $\frac{a}{r},a,ar$
$therefore\frac{a}{r},a,ar=56a(\frac{1}{r}+1+r)=56a\left(\frac{r^2+r+1}{r}\right)=56a=\frac{56r}{r^2+r+1}$
Given: $\frac{a}{r}-1,a-7,ar-21$ is an A.P.
$(a-7)-(\frac{a}{r}-1)=(ar-21)-(a-7)a-7-\frac{a}{r}+1=ar-21-a+7a(1-\frac{1}{r})-6=a(r-1)-14a(1-\frac{1}{r}-r+1)=-8a\left(\frac{2r-1-r^2}{r}\right)=-8\frac{56r}{r^2+r+1}\left(\frac{-r^2+2r-1}{r}\right)=-8$
$\Rightarrow \frac{7}{r^2+r+1}(-r^2+2r-1)=-1\Rightarrow -7r^2+14r-7=-r^2-r-114r+r=7r^2-r^2+7-115r=6r^2+6\Rightarrow 6r^2-15r+6=06r^2-3r-12r+6=03r(2r-1)-6(2r-1)=0(3r-6)(2r-1)=0r=\frac{6}{3}=2r=\frac{1}{2}a=\frac{56r}{r^2+r+1}a=\frac{56r}{r^2+r+1}=16=16$
The GP's are
$\frac{a}{r},a,ar\Rightarrow \frac{16}{2},16,16\times 2=8,16,32and\frac{16}{\frac{1}{2}},16,16\times \frac{1}{2}=32,16,8$