Question:
In an increasing, geometric series, the sum of the second and the sixth term is $\frac{25}{2}$ and the product of the third and fifth term is 25 . Then, the sum of $4^{\text {th }}, 6^{\text {th }}$ and $8^{\text {th }}$ terms is equal to :
Correct Option: 1
Solution:
$a r+a r^{5}=\frac{25}{2}$
$a r^{2} \times a r^{4}=25$
$a^{2} r^{6}=25$
$a r^{3}=5$
$a=\frac{5}{r^{3}}$
$\frac{5 r}{r^{3}}+\frac{5 r^{5}}{r^{3}}=\frac{25}{2}$
$\frac{1}{r^{2}}+r^{2}=\frac{5}{2}$
Put $r^{2}=t$
$\frac{t^{2}+1}{t}=\frac{5}{2}$
$2 t^{2}-5 t+2=0$
$2 t^{2}-4 t-t+2=0$
$(2 t-1)(t-2)=0$
$t=\frac{1}{2}, 2 \Rightarrow r^{2}=\frac{1}{2}, 2$
$r=\sqrt{2}$
$=a r^{3}+a r^{5}+a r^{7}$
$=a r^{3}\left(1+r^{2}+r^{4}\right)$
$=5[1+2+4]=35$