Question

Three numbers are in G.P, their product is 512. If we add 4 in first & second number, then they are in an A. P. then the sum of the original 3 numbers is

• A. 28
• B. 24
• C. 32
• D. 20

Answer 1

The correct option is A 28

Let the three numbers in G.P. be $a,ar,a{r}^2$.

Then,

$a\times ar\times a{r}^2=512$

$(ar{)}^3=512$

$ar=8$

Now,

$\frac{(a+4)+a{r}^2}{2}=ar+4$

$a+a{r}^2+4=2ar+8$

Putting $r=\frac{8}{a}$, we get,

$a+\frac{64}{a}=20$

$a^2-20a+64=0$

$a^2-16a-4a+64=0$

$a(a-16)-4(a-16)=0$

$(a-16)(a-4)=0$

$a=16,4$

Therefore, numbers are $4,8,16$ or $16,8,4$.

Sum of numbers = $4+8+16=28$.