Question

If $a,b,c$ be in G.P. & $lo{g}_{c}a,lo{g}_{b}c,lo{g}_{a}b$ be in A.P., then show that the common difference of the A.P is

• A. $\frac{1}{2}$
• B. $\frac{3}{2}$
• C. $\frac{5}{2}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $\frac{3}{2}$

Given $a,b,c$ are in GP and

$\log {}_{c}a,\log {}_{b}c,\log {}_{a}b$ are in AP

Let $r$ be the common ratio, then $b=a,c=a{r}^2\left(1\right)$
and $\log {}_{c}a,\log {}_{b}c,\log {}_{a}b$ are in AP

i.e. $\frac{\log a}{\log c},\frac{\log c}{\log b},\frac{\log b}{\log a}$ are in AP

From $\left(1\right)$ $\frac{\log a}{\log \left(a{r}^2\right)},\frac{\log \left(a{r}^2\right)}{\log \left(ar\right)},\frac{\log \left(ar\right)}{\log a}$ are in AP

$\frac{\log a}{\log a+2\log r},\frac{\log a+2\log r}{\log a+\log r},\frac{\log a+\log r}{\log a}$ are in AP

Put $\frac{\log r}{\log a}=x$ we get

$\frac{1}{1+2x},\frac{1+2x}{1+x},1+x$ are in AP

$\frac{2(1+2x)}{(1+x)}=(1+x)+\frac{1}{(1+2x)}2x^3-3x^2-3x=0x=\frac{1}{4}(3+\sqrt{3})\left(2\right)2d=(1+x)-\frac{1}{1+2x}$
From $\left(2\right)$ $2d=3\Rightarrow d=\frac{3}{2}$