### Sequences And Series Question 11

#### Question 11 - 30 January - Shift 2

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 abc is equal to

(1) 343

(2) 216

(3) $\frac{343}{8}$

(4) $\frac{125}{8}$

## Show Answer

#### Answer: (2)

#### Solution:

#### Formula: Sum of $\mathbf{n}$ terms of an AP, Geometric Means (Mean Proportional) (G.M.)(I)

As $a^{3}, b^{3}, c^{3}$ be in A.P. $\to a^{3}+c^{3}=2 b^{3}$

$\log _a^{b}, \log _c^{a}, \log _b^{c}$ are in G.P.

$\therefore \frac{\log b}{\log a} \cdot \frac{\log c}{\log b}=(\frac{\log a}{\log c})^{2}$

$\therefore(\log a)^{3}=(\log c)^{3} \Rightarrow a=c$

From (1) and (2)

$a=b=c$

$T_1=\frac{a+4 b+c}{3}=2 a ; d=\frac{a-8 b+c}{10}=\frac{-6 a}{10}=\frac{-3}{5} a$

$\therefore S _{20}=\frac{20}{2}[4 a+19(-\frac{3}{5} a)]$

$=10[\frac{20 a-57 a}{5}]$

$=-74 a$

$\therefore-74 a=-444 \Rightarrow a=6$

$\therefore abc=6^{3}=216$