Question

Three numbers form a GP. If the 3rd term is decreased by $64$, then the three numbers thus obtained will constitute an AP. If the second term of this AP is decreased by $8$, a GP will be formed again, then the numbers will be

• A. $4,20,36$
• B. $4,12,36$
• C. $4,20,100$
• D. none of these

Answer 1

The correct option is C $4,20,100$

$a,ar,a{r}^2\Rightarrow$ G.P

$a,ar,a{r}^2-64$ $\to$ AP

$\Rightarrow$ $2ar=a+a{r}^2-64$

$\Rightarrow$ $2r=1+r^2-\frac{64}{a}$

$\Rightarrow$ $\frac{64}{a}=r^2-2r+1$

$\Rightarrow$ $a=\frac{64}{r^2-2r+1}$

$a,ar-8,a{r}^2-64$ $\to$ G.P

$\Rightarrow$ ${(ar-8)}^2=(a{r}^2-64)a$

${(\frac{64r}{r^2-2r+1}-8)}^2=(\frac{64r^2}{r^2-2r+1}-64)\times \frac{64}{r^2-2r+1}$

$\Rightarrow 8^2{(\frac{8r}{r^2-2r+1}-1)}^2=64(\frac{r^2}{r^2-2r+1}-1)\times \frac{64}{r^2-2r+1}$

$\Rightarrow {\left(\frac{8r-r^2+2r-1}{r^2-2r+1}\right)}^2=\frac{(r^2-r^2+2r-1)}{r^2-2r+1}\frac{64}{r^2-2r+1}$

$\Rightarrow {(10r-r^2-1)}^2(2r-1)64$

$\Rightarrow 100r^2+r^4+1+2(-10r^3+r^2-10r)=64(2r-1)$

$\Rightarrow 100r^2+r^4+1-20r^3+2r^2-20r=128r-64$

$\Rightarrow r^4-20r^3+102r^2-148r+65=0$

$\Rightarrow {(r-1)}^2(r-5)(r-13)=0$

$\Rightarrow r=1,1,5,13$

$a(5,13)=\frac{64}{16},\frac{64}{12\times 12}=4,\frac{8}{9}$

with $a=4,r=5$

GP is $4,20,100$