Question

If $a,b,c$ be distinct positive integers in G .P and ${\log }_{c}a,{\log }_{b}c,{\log }_{a}b$ be in A.P., then common difference of this A.P. is-

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

Answer 1

Answer: (A)

$\frac{3}{2}$

Let r be the common ratio then $b=ar,c=a{r}^2$, and
$lo{g}_{c}a,lo{g}_{b}c,lo{g}_{a}b$ are in A.P.
$\Rightarrow \frac{lo{g}_{e}a}{lo{g}_{e}c},\frac{lo{g}_{e}c}{lo{g}_{e}b},\frac{lo{g}_{e}b}{lo{g}_{e}a}$ are in A.P.
$\Rightarrow \frac{lo{g}_{e}a}{lo{g}_e(ar{)}^2},\frac{lo{g}_3(ar{)}^2}{lo{g}_{e}ar},\frac{lo{g}_{e}ar}{lo{g}_{e}a}$ are in A.P.
so $\frac{lo{g}_{e}a}{lo{g}_{e}a+2lo{g}_{e}r},\frac{lo{g}_{e}a+2lo{g}_{e}r}{lo{g}_{e}a+lo{g}_{e}r},\frac{lo{g}_{e}a+lo{g}_{e}r}{lo{g}_{e}a}$ are in A.P.
Putting $\frac{lo{g}_{e}r}{lo{g}_{e}a}=x$
We get
$\frac{1}{1+2x},\frac{1+2x}{1+x},1+x$ are in A.P.
$\therefore \frac{2(1+2x)}{(1+x)}=(1+x)+\frac{1}{(1+2x)}$
$\Rightarrow 2x^3-3x^2-3x=0$
Since a, b, c are distinct so $r\ne 1$, so $x\ne 0$
$\therefore 2x^2-3x-3=0$
$x=\frac{1}{4}(3\pm \sqrt{33})$ then $(1+x)-\frac{1}{(1+2x)}=3$, so the common differences of A.P. is 3/2