Let a,b,c > 1,a^3,b^3 and c^3 be in A.P., and _ab,_ca and _bc 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

As a^3, b^3, c^3 be in A.P.\\ a^3+c^3=2 b^3 \\ _a^b, _c^a, _b^c are in G.P.\\ b a c b \\=( a c)^2 \\ ( a)^3=( c)^3 \\ a=c . .(2)\\ F

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 + 4b + c}{3} and the common difference is \frac{a - 8b + c}{10} is -444 , then abc is equal to: 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

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

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