Let a, b, c1, 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 :

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 :

Accepted Answer

Answer: B. 216 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

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 :

Answer

Explanation

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