Let a,b,c > 1,a^3,b^3 and c^3 be in A.P., and log _ab,log _ca and log _bc be in G.P. If the sum of first 20 terms of an A.P., whose first term is (a + 4b + c / 3) and the common difference is (a - 8b + c / 10) is -444 , then abc is equal to

As a^3, b^3, c^3 be in A.P.\ -> a^3+c^3=2 b^3 \ log _a^b, log _c^a, log _b^c are in G.P.\ Therefore (log b / log a) * (log c / log b) =((log a / log c))^2 \ Therefore (log a)^3=(log c)^3 \ => a=c ... . .(2) From (1) and (2)\ a=b=c Gamma_1=(a+4 b+c / 3)=2 a \ d=(a-8 b+c / 10)=(-6 a / 10)=(-3 / 5) a \ Therefore S_20=(20 / 2)[4 a+19(-(3 / 5) a)] \ =10[(20 a-57 a / 5)] \ =-74 a \ Therefore -74 a=-444 => a=6 \ Therefore a b c=6^3=216

Let a,b,c > 1,a^{3},b^{3} and c^{3} be in A.P., and \... | Sequence & Series (Mathematics)

As $a^3, b^3, c^3$ be in A.P. $\\ \rightarrow a^3+c^3=2 b^3$ $\\ \log _a^b, \log _c^a, \log _b^c$ are in G.P. $\\ \therefore \frac{\log b}{\log a} \cdot \frac{\log c}{\log b} \\=\left(\frac{\log a}{\log c}\right)^2 \\ \therefore(\log a)^3=(\log c)^3 \\ \Rightarrow a=c \ldots . .(2)\\$ From (1) and (2) $\\ a=b=c \Gamma_1=\frac{a+4 b+c}{3}=2 a \\ d=\frac{a-8 b+c}{10}=\frac{-6 a}{10}=\frac{-3}{5} a \\ \therefore S_{20}=\frac{20}{2}\left[4 a+19\left(-\frac{3}{5} a\right)\right] \\ =10\left[\frac{20 a-57 a}{5}\right] \\ =-74 a \\ \therefore-74 a=-444 \Rightarrow a=6 \\ \therefore a b c=6^3=216$

Explanation

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