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 a + 4 b + c 3 and the common difference is a - 8 b + c 10 is - 444 , then a b c is eq

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 [2023]

Smart Achievers · Accepted Answer

(2)

$a^{3}, b^{3}, c^{3} are in A.P. and \log_{a} b, \log_{c} a, \log_{b} c are in G.P.$

$\frac{\log_{c} a}{\log_{a} b} = \frac{\log_{b} c}{\log_{c} a}$

$\Rightarrow \frac{\log_{e} a \times \log_{e} a}{\log_{e} c \times \log_{e} b} = \frac{\log_{e} c}{\log_{e} b} \times \frac{\log_{e} c}{\log_{e} a}$

$\Rightarrow {(\log_{e} a)}^{3} = {(\log_{e} c)}^{3} \Rightarrow \log_{e} a = \log_{e} c \Rightarrow a = c ...(i)$

$\Rightarrow b^{3} - a^{3} = c^{3} - b^{3} \Rightarrow b^{3} - a^{3} = a^{3} - b^{3}$

$\Rightarrow b^{3} = a^{3} \Rightarrow a = b \Rightarrow a = b = c$

$First term, a_{1} = \frac{a + 4 b + c}{3}, d = \frac{a - 8 b + c}{10}$

$S_{20} = - 444$

$\Rightarrow - 444 = 10 [2 \times \frac{6 a}{3} + 19 \times (\frac{- 6 a}{10})]$

$\Rightarrow - 444 = 10 [4 a - \frac{57}{5} a] \Rightarrow - 444 = 10 [\frac{20 a - 57 a}{5}]$

$\Rightarrow - 444 = 2 (- 37 a) \Rightarrow - 222 = - 37 a$

$\Rightarrow a = 6$ $∴$ $a b c = a^{3} = 6^{3} = 216$

Solution

Q.
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 [2023]

1 343

2 216

3 $\frac{343}{8}$

4 $\frac{125}{8}$

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