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Explanation:

y = $\frac{1}{2+{\displaystyle \frac{1}{3+{\displaystyle \frac{1}{2+{\displaystyle \frac{1}{3+....}}}}}}}$

y = $\frac{1}{2+{\displaystyle \frac{1}{3+y}}}$

y = $\frac{3+y}{2y+7}$

∴ 2y² + 7y = 3 + y

∴ 2y² + 6y – 3 = 0

∴ y = $\frac{-6\pm \sqrt{36-4(-3)\left(2\right)}}{4}$ = $\frac{\pm \sqrt{15}-3}{2}$

As all the terms in y are positive, y is positive.

∴ y = $\frac{\sqrt{15}-3}{2}$

Hence, option (d).