Question

Let $a,b,c>1,a^3,b^3$ and $c^3$ be in A.P., and ${\log }_{a}b,{\log }_{c}a$ and ${\log }_{b}c$ be in G.P. If the sum of first $20$ terms of an A.P., whose first term is $\frac{a+4b+c}{3}$ and the common difference is $\frac{a-8b+c}{10}$ is $-444$, then $abc$ is equal to:

• A. 343
• B. 216
• C. $\frac{343}{8}$
• D. $\frac{125}{8}$

Answer 1

Answer: (B)

As $a^3,b^3,c^3$ be in A.P. $\to a^3+c^3=2b^3$.... (1)
${\log }_a^b,{\log }_c^a,{\log }_b^c$ are in G.P.
$\therefore \frac{\log b}{\log a}\cdot \frac{\log c}{\log b}={\left(\frac{\log a}{\log c}\right)}^2$
$\therefore (\log a{)}^3=(\log c{)}^3\Rightarrow a=c$......(2)
From (1) and (2)
$a=b=c$
$T_1=\frac{a+4b+c}{3}=2a;d=\frac{a-8b+c}{10}=\frac{-6a}{10}=\frac{-3}{5}a$
$\therefore S_{20}=\frac{20}{2}\left[4a+19\left(-\frac{3}{5}a\right)\right]$
$=10\left[\frac{20a-57a}{5}\right]$
$=-74a$
$\therefore -74a=-444\Rightarrow a=6$
$\therefore abc=6^3=216$