Let a, b, c > 1, a3, b3 and c3 be in A.P., and logab, logca and logbc be in G.P. If the sum of first 20 terms of an A.P., whose first term is a+4b+c3 and the common difference is a−8b+c10 is -444, then abc is equal to:

a3, b3, c3 are in A.P ⇒2b3 = a3 + c3logab,logca,logbc⁡ in U.P ⇒logca2=logab⋅logbc=logac∴logca3=1⇒a=c2b3=a3+c3⇒a=b=cA=a+4b+c3=2a,D=a−8b+c10=−3a5S20=2024a+19−3a5=−444⇒a=6∴abc=a3=216

Let a, b, c > 1, a3, b3 and c3 be in A.P., and logab, logca and logbc be in G.P. If the sum of first 20 terms of an A.P., whose first term is a+4b+c3 and the common difference is a−8b+c10 is -444, then abc is equal to:

a3, b3, c3 are in A.P ⇒2b3 = a3 + c3logab,logca,logbc⁡ in U.P ⇒logca2=logab⋅logbc=logac∴logca3=1⇒a=c2b3=a3+c3⇒a=b=cA=a+4b+c3=2a,D=a−8b+c10=−3a5S20=2024a+19−3a5=−444⇒a=6∴abc=a3=216

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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:

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343

216

343/8

125/8

answer is B.

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Detailed Solution

a³, b³, c³ are in A.P $\Rightarrow$ 2b³ = a³ + c³
$\begin{array}{l} \log_{a}^{b}, \log_{c}^{a}, \log_{b}^{c} in U.P \Rightarrow {(\log_{c}^{a})}^{2} = \log_{a}^{b} \cdot \log_{b}^{c} = \log_{a}^{c} \\ ∴ {(\log_{c}^{a})}^{3} = 1 \Rightarrow a = c \\ 2 b^{3} = a^{3} + c^{3} \Rightarrow a = b = c \\ A = \frac{a + 4 b + c}{3} = 2 a, D = \frac{a - 8 b + c}{10} = \frac{- 3 a}{5} \\ S_{20} = \frac{20}{2} \{4 a + 19 (\frac{- 3 a}{5})\} = - 444 \\ \Rightarrow a = 6 ∴ abc = a^{3} = 216 \end{array}$