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Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that `P^2R^n=S^n`. |
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Answer» Let the three terms of the given GP be `a/r`, a and ar. Then, `S=(a/r+a+ar)=a((1+r+r^(2))/r), P=(a/rxxaxxar)=a^(3)` and `R=(r/a+1/a+1/(ar))=1/a(r+1+1/r)=1/a((r^(2)+r+1)/r)=1/a((1+r+r^(2))/r)`. `:. P^(2)R^(3)=a^(6) xx((1+r+r^(2))^(3))/(a^(3)r^(3))=a^(3)((1+r+r^(2))/r)^(3)=S^(3)` Hence, `P^(2)R^(3)=S^(3)`. |
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