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Find the value of n so that `(a^(n+1)+b^(n+1))/(a^n+b^n)`may be the geometric mean between a and b. |
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Answer» As given expession is geometric mean of `a` and `b`. `(a^(n+1)+b^(n+1))/(a^n+b^n) = (ab)^(1/2)` `=>(a^(n+1)+b^(n+1)) = a^(n+1/2)b^(1/2)+b^(n+1/2)a^(1/2)` `=>a^(n+1)-a^(n+1/2)b^(1/2) = b^(n+1/2)a^(1/2)-b^(n+1)` `=>a^(n+1/2)(a^(1/2)-b^(1/2)) = b^(n+1/2)(a^(1/2)-b^(1/2))` `=>(a/b)^(n+1/2) = 1` As `a !=b`, `:. n+1/2 = 0` `n = -1/2` |
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