Question

Let three positive numbers $a,b,c$ be in $GP$ such that $a,b+8,c$ are in $AP$, while $a,b,+8,c+64$ are in $GP$. Then the $AM$ of $a,b,c$ is

• A. $\frac{52}{3}$
• B. $\frac{28}{9}$
• C. $26$
• D. $14$

Answer 1

The correct option is A $\frac{52}{3}$

Given

$b^2=ac$ ___(1)

$2(b+8)=a+c$ ____(2)

$(b+8{)}^2=a(c+64)$ ___(3)

$\Rightarrow (b+8{)}^2=ac+64a=b^2+64a$

$\Rightarrow 8(2b+8)=64a$

$\Rightarrow b+4=4a$ ____(4)

From (2) and (4)

$2(4a+4)=a+c$

$\Rightarrow 7a=c-8$
From (1),

$(4a-4{)}^2=a(7a+8)$

$\Rightarrow 9a^2-40a+16=0$

$\Rightarrow a=\frac{4}{9},4$
Correspondingly, the triplets $\{a,b,c\}$ are $\{\frac{4}{9},\frac{-20}{9},\frac{100}{9}\}$ and $\{4,12,36\}$

But $a,b,c$ are all positive

$\therefore AM=\frac{4+12+36}{3}=\frac{52}{3}$