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

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