Let a, b, c1, a^3, b^3 and c^3 be in A.P., and _a b, _c a and _b c be in G.P.

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

$2{b^3} = {a^3} + {c^3}$

$${\left( {{{\log a} \over {\log c}}} \right)^2} = \left( {{{\log b} \over {\log a}}} \right)\left( {{{\log c} \over {\log b}}} \right)$$

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

$$ \Rightarrow \log a = \log c$$

$$ \Rightarrow a = c$$

$$ \Rightarrow a = b = c$$

$${T_1} = 2a,d =  - {{3a} \over 5}$$

${S_{20}} =  - 444$

$$ \Rightarrow {{20} \over 2}\left( {2(2a) + (19)\left( { - {{3a} \over 5}} \right)} \right) =  - 444$$

$$ \Rightarrow 10{{(20a - 57a)} \over 5} =  - 444$$

$$ \Rightarrow 37a = 222$$

$$ \Rightarrow a = 6$$

$$ \Rightarrow abc = {(6)^3} = 216$$

Solutions · 1