Question

Three numbers $a,b$ and $c$ are in geometric progression. If $4a,5b$ and $4c$ are in arithmetic progression and $a+b+c=70$ , then the value of $\left|c-a\right|$ is equal to

• A. $10$
• B. $20$
• C. $30$
• D. $40$

Answer 1

Answer: (C)

$a+b+c=70$
$\Rightarrow a+ar+a{r}^2=70$
Also, $10b=4a+4c$
$\Rightarrow 10ar=4a+4a{r}^2$
$\Rightarrow 2r^2-5r+2=0\Rightarrow r=2,\frac{1}{2}$
For $r=2$ , we have,
$a\left(1+r+r^2\right)=70\Rightarrow a=\frac{70}{1+r+r^2}=\frac{70}{1+2+4}=10$
$\Rightarrow a=10,b=20,c=40\Rightarrow \left|c-a\right|=30$
For $r=\frac{1}{2}$ , we have,
$a=\frac{70}{1+r+r^2}=\frac{70}{1+\frac{1}{2}+\frac{1}{4}}=\frac{70\times 4}{4+2+1}=40$
$\Rightarrow b=20,c=10$
$\Rightarrow \left|c-a\right|=\left|10-40\right|=30$
So, in both the cases $\left|c-a\right|=30$