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(A) Let the three successive terms of the $G.P.$ be $a, ar, ar^2$ where $a > 0$ and $r > 1$.For these to be sides of a triangle,the sum of any two sid

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(A) Let the three successive terms of the $G.P.$ be $a, ar, ar^2$ where $a > 0$ and $r > 1$.For these to be sides of a triangle,the sum of any two sides must be greater than the third side.$a + ar > ar^2 \implies 1 + r > r^2 \implies r^2 - r - 1 The roots of $r^2 - r - 1 = 0$ are $r = \frac{1 \pm \sqrt{5}}{2}$.Since $r > 1$,the condition $r^2 - r - 1 Given $\sqrt{5} \approx 2.236$,we have $\frac{1 + 2.236}{2} = 1.618$.So,$1 Then,$[r] = 1$.For $[-r]$,since $1 The greatest integer less than or equal to $-r$ is $[-r] = -2$.Therefore,$3[r] + [-r] = 3(1) + (-2) = 3 - 2 = 1$.

If three successive terms of a $G.P.$ with common ratio $r$ $(r > 1)$ are the lengths of the sides of a triangle,and $[r]$ denotes the greatest integer less than or equal to $r$,then $3[r] + [-r]$ is equal to:

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