The number of 5-tuples (a, b, c, d, e) of pos | vedclass

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(B) The sum of the interior angles of a convex pentagon is $(5-2) 	imes 180^{\circ} = 540^{\circ}$.Let the angles be $a, a+d, a+2d, a+3d, a+4d$ where

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$36$

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(B) The sum of the interior angles of a convex pentagon is $(5-2) 	imes 180^{\circ} = 540^{\circ}$.Let the angles be $a, a+d, a+2d, a+3d, a+4d$ where $d \geq 0$.The sum is $5a + 10d = 540^{\circ}$,which simplifies to $a + 2d = 108^{\circ}$.Since $a$ and $d$ are integers and $a > 0$,we have $2d = 108 - a$.For the pentagon to be convex,each angle must be less than $180^{\circ}$. The largest angle is $e = a + 4d$.Substituting $a = 108 - 2d$,we get $e = (108 - 2d) + 4d = 108 + 2d$.We require $108 + 2d Also,since $a$ is a positive integer,$a = 108 - 2d > 0$,so $2d Since $d$ must be a non-negative integer,$d$ can take values from $0, 1, 2, \dots, 35$.This gives a total of $36$ possible values for $d$,and each $d$ uniquely determines $a$.

The number of $5$-tuples $(a, b, c, d, e)$ of positive integers such that:
$I.$ $a, b, c, d, e$ are the measures of angles of a convex pentagon in degrees.
$II.$ $a \leq b \leq c \leq d \leq e$.
$III.$ $a, b, c, d, e$ are in an arithmetic progression.

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The number of $5$-tuples $(a, b, c, d, e)$ of positive integers such that:$I.$ $a, b, c, d, e$ are the measures of angles of a convex pentagon in degrees.$II.$ $a \leq b \leq c \leq d \leq e$.$III.$ $a, b, c, d, e$ are in an arithmetic progression.