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If a, b, c and d are in G.P. show that `(a^2+b^2+c^2)(b^2+c^2+d^2)=(a b+b c+c d)^2`.

Answer» Let r be the common ratio of the GP a, b, c, d.
Then, `b=ar, c=ar^(2)` and `d=ar^(3)`.
`:. LHS=(a^(2)+b^(2)+c^(2))(b^(2)+c^(2)+d^(2))`
`=(a^(2)+a^(2)r^(2)+a^(2)r^(4))(a^(2)r^(2)+a^(2)r^(4)+a^(2)r^(6))`
`=a^(4)r^(2) (1+r^(2)+r^(4))^(2)`.
And, `RHS =(ab+bc+cd)^(2)=(a^(2)r+a^(2)r^(3)+a^(2)r^(5))^(2)`
`=a^(4)r^(2) (1+r^(2)+r^(4))^(2)`.
Hence, `(a^(2)+b^(2)+c^(2)) (b^(2)+c^(2)+d^(2))=(ab+bc+cd)^(2)`.


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