Question

If $a,b,c$ are in A.P. and $a,c-b,b-a$ are in G.P., where$a\ne b\ne c$, then $a:b:c$ is ___.

• A. 1:3:5
• B. 1:2:4
• C. 1:2:3
• D. 3:2:1

Answer 1

The correct option is C

1:2:3

Since $a,b,c$ are in A.P., we have
$b=\frac{1}{2}(a+c)...\left(1\right)$
Since $a,c-b,b-a$ are in G.P., we have
${(c-b)}^2=a(b-a)...\left(2\right)$
Substituting the value of $b$ from $\left(1\right)$ in $\left(2\right),$ we get
${[c-\frac{1}{2}(a+c\left)\right]}^2=a\left[\frac{1}{2}\right(a+c)-a].$
$\Rightarrow \frac{(c-a{)}^2}{4}=\frac{a(c-a)}{2}$
$\Rightarrow (c-a{)}^2=2a(c-a)$
$\Rightarrow c-a=2a(\because c\ne a)$
$\Rightarrow c=3a\Rightarrow b=2a\left(\text{from}\right(1\left)\right)$

∴ $\frac{a}{1}=\frac{b}{2}=\frac{c}{3}$
i.e., $a:b:c=1:2:3$