The lengths of the sides of a triangle are successive terms of a geometric progression. Let $A$ and $C$ be the smallest and the largest interior angles of the triangle respectively. If the shortest side has length $16$ centimeters and $\frac{sin A - 2 sin B + 3 sin C}{sin C - 2 sin B + 3 sin A} = \frac{19}{9},$ then the perimeter of the triangle in centimeters is
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Solution

Let the lengths of the sides of the triangle in centimetres be $16, 16 r, 16 r^{2}$ ( $r > 1$ )
Then, $\frac{1 - 2 r + 3 r^{2}}{r^{2} - 2 r + 3} = \frac{19}{9}$
$\Rightarrow 9 - 18 r + 27 r^{2} = 19 r^{2} - 38 r + 57$
$\Rightarrow 8 r^{2} + 20 r - 48 = 0$
$\Rightarrow 2 r^{2} + 5 r - 12 = 0$
$\Rightarrow (r + 4) (2 r - 3) = 0$
$\Rightarrow r = \frac{3}{2}$
Hence, the perimeter of the triangle is $16 (1 + \frac{3}{2} + \frac{9}{4}) = 76$ cm

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