If the sum of the second, fourth and sixth terms of a G.P. of positive terms is \(21\) and the sum of its eighth, tenth …

Let the first term of the G.P. be \(a\) and the common ratio be \(r\).

Given

\(a r+a r^3+a r^5=21 \Rightarrow a r\left(1+r^2+r^4\right)=21 \ldots (1)\)

and \(a r^7+a r^9+a r^{11}=15309 \Rightarrow a r^7\left(1+r^2+r^4\right)=15309 \ldots (2)\)

Divide Equation 2 by Equation 1

\(\frac{a r^7\left(1+r^2+r^4\right)}{a r\left(1+r^2+r^4\right)}=\frac{15309}{21}\)

\(r^6=729\)

\(r=3\)

from eq (1)

\(\begin{aligned} & 3 a(1+9+81)=21 \\ & 3 a(91)=21 \Rightarrow 273 a=21 \\ & a=\frac{21}{273}=\frac{1}{13}\end{aligned}\)

Now

\(\begin{aligned} & S_9=\frac{\frac{1}{13}\left(3^9-1\right)}{3-1} \\ & S_9=\frac{1}{13} \cdot \frac{19683-1}{2} \\ & S_9=\frac{19682}{26} \\ & S_9=757\end{aligned}\)