Question

Let three positive numbers $a,b,c$ are in geometric progression, such that $a,b+8,c$ are in arithmetic progression and $a,b+8,c+64$ are in geometric progression. If the arithmetic mean of $a,b,c$ is $k$ , then $\frac{3}{13}k$ is equal to

Answer 1

Answer: 4

Here, $b^2=ac,2\left(b+8\right)=a+c,$
${\left(b+8\right)}^2=a\left(c+64\right)$
$\Rightarrow b^2+64+16b=ac+64a$
$\Rightarrow b+4=4a\Rightarrow b=4a-4$
$2\left(b+8\right)=a+c\Rightarrow 2\left(4a+4\right)=a+c$
$\Rightarrow c=7a+8$
$b^2=ac\Rightarrow {\left(4a-4\right)}^2=a\left(7a+8\right)$
$\Rightarrow 16a^2+16-32a=7a^2+8a$
$\Rightarrow 9a^2-40a+16=0$
$\Rightarrow \left(9a-4\right)\left(a-4\right)=0\Rightarrow a=\frac{4}{9},4$
But for $a=\frac{4}{9}\Rightarrow b<0$
So, $a=4\Rightarrow b=12$ and $c=36$
$\Rightarrow k=\frac{a+b+c}{3}=\frac{52}{3}$
$\Rightarrow \frac{3k}{13}=\frac{3}{13}\times \frac{52}{3}=4$