Question

Sum of nine terms of the series log $a+\log \frac{a^2}{b}+\log \frac{a^3}{b^2}+\log \frac{a^4}{b^3}+.....is-$

• A. $36\log a-28\log b$
• B. $45\log a-36\log b$
• C. $55\log a-45\log b$
• D. $66\log a-55\log b$

Answer 1

The correct option is B $45\log a-36\log b$

Solution :-

Sum of nine terms

$loga+log\frac{a^2}{b^2}+log\frac{a^3}{b^3}+log\frac{a^2}{b^4}+...$

$=log[a\times \frac{a^2}{b}\times \frac{a^3}{b^2}\times \frac{a^4}{b^3}\times \frac{a^5}{b^4}\times \frac{a^6}{b^5}\times \frac{a^7}{b^6}\times \frac{a^8}{b^7}\times \frac{a^9}{b^8}]$ $[\because \log x+\log y=\log xy]$

$=log\left[\frac{a^{1+2+3+4+5+6+7+8+9}}{b^{1+2+3+4+5+6+7+8}}\right]$

$=log\left[\frac{a^{45}}{b^{36}}\right]$

$=log\left(a^{45}\right)-log\left(b^{36}\right)$

$=45loga-36logb$