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 $\frac{a+4 b+c}{3}$ and the common difference is $\frac{a-8 b+c}{10}$ is $-444$, then $a b c$ 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. $\rightarrow a ^3+ c ^3=2 b ^3$.... (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 $
$T _1=\frac{ a +4 b + c }{3}=2 a ; d =\frac{ a -8 b + c }{10}=\frac{-6 a }{10}=\frac{-3}{5} a$
$ \therefore S _{20}=\frac{20}{2}\left[4 a +19\left(-\frac{3}{5} a \right)\right]$
$=10\left[\frac{20 a -57 a }{5}\right]$
$ =-74 a $
$ \therefore-74 a =-444 \Rightarrow a =6$
$ \therefore a b c=6^3=216$