JEE MAIN - Mathematics (2023 - 30th January Evening Shift - No. 5)
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 :
343
216
$\frac{343}{8}$
$\frac{125}{8}$
Explanation
$$2{b^3} = {a^3} + {c^3}$$
$${\left( {{{\log a} \over {\log c}}} \right)^2} = \left( {{{\log b} \over {\log a}}} \right)\left( {{{\log c} \over {\log b}}} \right)$$
$$ \Rightarrow {(\log a)^3} = {(\log c)^3}$$
$$ \Rightarrow \log a = \log c$$
$$ \Rightarrow a = c$$
$$ \Rightarrow a = b = c$$
$${T_1} = 2a,d = - {{3a} \over 5}$$
$${S_{20}} = - 444$$
$$ \Rightarrow {{20} \over 2}\left( {2(2a) + (19)\left( { - {{3a} \over 5}} \right)} \right) = - 444$$
$$ \Rightarrow 10{{(20a - 57a)} \over 5} = - 444$$
$$ \Rightarrow 37a = 222$$
$$ \Rightarrow a = 6$$
$$ \Rightarrow abc = {(6)^3} = 216$$
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