Question

The sequence ${\log }_{12}162,{\log }_{12}x,{\log }_{12}y,{\log }_{12}z,{\log }_{12}1250$ is an arithmetic progression. The value of $x$ is

• A. $125\sqrt{3}$
• B. 270
• C. $162\sqrt{5}$
• D. 434

Answer 1

Answer: (B)

Let the terms are $a,a+d,a+2d,a+3d,a+4d$
$\Rightarrow 4d={\log }_{12}1250-{\log }_{12}162$
$\therefore 4d={\log }_{12}\left(\frac{1250}{162}\right)$
$={\log }_{12}\left(\frac{625}{81}\right)$
$={\log }_{12}{\left(\frac{5}{3}\right)}^4$
$=4{\log }_{12}\left(\frac{5}{3}\right)$
$d={\log }_{12}\left(\frac{5}{3}\right)$
Now, ${\log }_{12}x-{\log }_{12}162={\log }_{12}\left(\frac{5}{3}\right)$
${\log }_{12}x={\log }_{12}\left(102\times \frac{5}{3}\right)={\log }_{12}270$
$\Rightarrow x=270$