If a_{1} (0), a_{2}, a_{3}, a_{4}, a_{5} are | vedclass

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(D) Let the terms of the $G$.$P$. be $a, ar, ar^{2}, ar^{3}, ar^{4}$.Given $3a_{2} + a_{3} = 2a_{4}$,we have $3ar + ar^{2} = 2ar^{3}$.Since $a_{1} > 0

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$40$

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(D) Let the terms of the $G$.$P$. be $a, ar, ar^{2}, ar^{3}, ar^{4}$.Given $3a_{2} + a_{3} = 2a_{4}$,we have $3ar + ar^{2} = 2ar^{3}$.Since $a_{1} > 0$,$a 
eq 0$,so $3 + r = 2r^{2}$,which implies $2r^{2} - r - 3 = 0$.Solving the quadratic equation: $(2r - 3)(r + 1) = 0$,so $r = \frac{3}{2}$ or $r = -1$.Given $a_{2} + a_{4} = 2a_{3} + 1$,we have $ar + ar^{3} = 2ar^{2} + 1$,or $a(r + r^{3} - 2r^{2}) = 1$.If $r = -1$,$a(-1 - 1 - 2) = 1 \implies -4a = 1 \implies a = -\frac{1}{4}$. Since $a_{1} > 0$,this is rejected.If $r = \frac{3}{2}$,$a(\frac{3}{2} + \frac{27}{8} - 2(\frac{9}{4})) = 1 \implies a(\frac{12 + 27 - 36}{8}) = 1 \implies a(\frac{3}{8}) = 1 \implies a = \frac{8}{3}$.Now,$a_{2} + a_{4} + 2a_{5} = ar + ar^{3} + 2ar^{4} = a(r + r^{3} + 2r^{4})$.Substituting $a = \frac{8}{3}$ and $r = \frac{3}{2}$:$= \frac{8}{3} (\frac{3}{2} + \frac{27}{8} + 2 	imes \frac{81}{16}) = \frac{8}{3} (\frac{3}{2} + \frac{27}{8} + \frac{81}{8}) = \frac{8}{3} (\frac{12 + 27 + 81}{8}) = \frac{8}{3} (\frac{120}{8}) = \frac{8}{3} 	imes 15 = 40$.

If $a_{1} (>0), a_{2}, a_{3}, a_{4}, a_{5}$ are in a $G$.$P$.,$a_{2} + a_{4} = 2a_{3} + 1$ and $3a_{2} + a_{3} = 2a_{4}$,then $a_{2} + a_{4} + 2a_{5}$ is equal to

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