Question

Find the sum of n terms of the series $\log a+\log \frac{a^2}{b}+\log \frac{a^3}{b^2}+\log \frac{a^4}{b^3}+$.... to n terms .

• A. $\frac{n}{4}[n\log \frac{b}{a}+\log ab]$
• B. $\frac{n}{4}[n\log \frac{a}{b}+\log ab]$
• C. $\frac{n}{2}[n\log \frac{a}{b}+\log ab]$
• D. $\frac{n}{8}[n\log \frac{b}{a}+\log ab]$

Answer 1

The correct option is C $\frac{n}{2}[n\log \frac{a}{b}+\log ab]$

$T_n-T_{n-1}=\log \frac{a^n}{b^{n-1}}-\log \frac{a^{n-1}}{b^{n-2}}$

$\log \frac{a^n}{b^{n-1}}\cdot \frac{b^{n-2}}{a^{n-1}}=\log \frac{a}{b}=$ constant as it is independent of n.

Hence the given series is an A.P. whose $A=\log a$ and $D=\log \frac{a}{b}$

$\therefore S=\frac{n}{2}[2\log a+[n-1]\log \frac{a}{b}]$ $=\frac{n}{2}[n\log \frac{a}{b}+2\log a-(\log a-\log b)]$
$=\frac{n}{2}[n\log \frac{a}{b}+\log ab]$

Hence, option C.