Question

The value of $\sum _{r=1}^{n}log\left(\frac{a^r}{b^{r-1}}\right)$ is

• A. $\frac{n}{2}log\left(\frac{a^n}{b^n}\right)$
• B. $\frac{n}{2}log\left(\frac{a^{n+1}}{b^n}\right)$
• C. $\frac{n}{2}log\left(\frac{a^{n+1}}{b^{n-1}}\right)$
• D. $\frac{n}{2}log\left(\frac{a^{n+1}}{b^{n+1}}\right)$

Answer 1

The correct option is C $\frac{n}{2}log\left(\frac{a^{n+1}}{b^{n-1}}\right)$

The given series is
$loga+log\left(\frac{a^2}{b}\right)+log\left(\frac{a^3}{b^2}\right)+log\left(\frac{a^4}{b^3}\right)+............+log\left(\frac{a^n}{b^{n-1}}\right)$
This is an A.P. with first term loga and the common difference
$log\left(\frac{a^2}{b}\right)-loga=log\left(\frac{a}{b}\right)$
Therefore the sum of n terms is
$\frac{n}{2}[loga+log(\frac{a^n}{b^{n-1}}\left)\right]=\frac{n}{2}log\left(\frac{a^{n+1}}{b^{n-1}}\right)$
​ Trick : Check for $n=1,2.$