8. sequence and series — MATHS STD 11 Science — Question

$3 a _{2}+ a _{3}=2 a _{4}$

$3 ar + ar ^{2}=2 ar ^{3}$

$3+ r =2 r ^{2}$

$2 r ^{2}- r -3=0$

$r =-1$ and $r =\frac{3}{2}$

$a _{2}+ a _{4}=2 a _{3}+1$

$ar + ar ^{3}=2 ar ^{2}+1$

$a \left( r + r ^{3}-2 r ^{2}\right)=1$

$a\left(\frac{3}{2}+\frac{27}{8}-\frac{18}{4}\right)=1$

$a=\frac{8}{3}$

$When \;r =-1, a =-\frac{1}{4}\; (rejected, a _{1} > 0)$

$r =\frac{2}{3}, a =\frac{8}{3}(\text { selected })$

Now

$a_{2}+a_{4}+2 a_{5}$

$=\frac{8}{3} \times \frac{3}{2}+\frac{8}{3} \times \frac{27}{8}+2 \times \frac{8}{3} \times \frac{81}{16}$

$=4+9+27=40$