Let G.P. be a₁ = a, a₂ = ar, a₃ = ar², .........

$$\because$$ 3a₂ + a₃ = 2a₄

$$\Rightarrow$$ 3ar + ar² = 2ar³

$$\Rightarrow$$ 2ar² $$-$$ r $$-$$ 3 = 0

$$\therefore$$ r = $$-$$1 or $${3 \over 2}$$

$$\because$$ a₁ = a > 0 then r $$\ne$$ $$-$$1

Now, a₂ + a₄ = 2a₃ + 1

ar + ar³ = 2ar² + 1

$$a\left( {{3 \over 2} + {{27} \over 8} - {9 \over 2}} \right) = 1$$

$$\therefore$$ a = $${8 \over 3}$$

$$\therefore$$ a₂ + a₄ + 2a₅ = a(r + r³ + 2r⁴)

$$ = {8 \over 3}\left( {{3 \over 2} + {{27} \over 8} + {{81} \over 8}} \right) = 40$$