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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:

Option: 1

\frac{125}{8}


Option: 2

216


Option: 3

343


Option: 4

\frac{343}{8}


Answers (1)

If \log _{\mathrm{a}} \mathrm{b}, \log _{\mathrm{c}} a, \log _{b} c \rightarrow G.P.

\left(\log _{c} a\right)^{2}=\log _{a} b \times \log _{b} c
\left(\log _{\mathrm{c}} \mathrm{a}\right)^{2}=\log _{\mathrm{a}} \mathrm{c}

\Rightarrow\left(\log _{c} a\right)^{2}=\frac{1}{\log _{c} a}
\Rightarrow\left(\log _{\mathrm{c}} \mathrm{a}\right)^{3}=1
\Rightarrow \log _{\mathrm{c}} \mathrm{a}=1
\mathrm{a}=\mathrm{c}

\text { If } a^{3} b^{3} c^{3} \rightarrow \text { A.P }

2 b^{3}=a^{3}+c^{3}
\text { If } \mathrm{a}=\mathrm{c}
\Rightarrow \mathrm{a}=\mathrm{b}=\mathrm{c}

\text { For AP }
A=\frac{a+4 a+a}{3} \quad D=\frac{a-8 a+a}{10}
\mathrm{~A}=2 \mathrm{a} \quad \mathrm{D}=\frac{-3 \mathrm{a}}{5}

S_{20}=\frac{20}{2}\left[2 \times 2 a+(20-1)\left(\frac{-3 a}{5}\right)\right]
        =10\left[4 a-\frac{57 a}{5}\right]
      =10\left[-\frac{37 a}{5}\right]=-444
\Rightarrow \mathrm{a}=\frac{444 \times 5}{37 \times 10}
\mathrm{a}=6
\Rightarrow \mathrm{abc}=6 \times 6 \times 6=216
 

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Kshitij

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