R S Aggarwal Solutions for Exercise 1: EXERCISE 12A

Which term of the $\mathrm{GP} \sqrt{3},3,3\sqrt{3}\dots$ is $729 ?$

Find the geometric series whose $5^{th}$ and $8^{th}$ terms are $80$ and $640$ respectively.

Find the $GP$ whose $4^{th}$ and $7^{th}$ terms are $\frac{1}{18}$ and $\frac{-1}{486}$ respectively.

The $5^{th},8^{th}$ and ${11}^{th}$ terms of a $GP$ are $a,b,c$ respectively. Show that $b^2=ac$.

The first term of a $GP$ is $-3$ and the square of the second term is equal to its $4^{th}$ term. Find its $7^{th}$ term.

Find the value of $k$, if the $6^{th}$ term from the end of $GP 8,4,2\dots \frac{1}{1024}$ is $\frac{1}{k}$

Find the $4^{\mathrm{th}}$ term from the end of the GP $\frac{2}{27},\frac{2}{9},\frac{2}{3},\dots ,162$.

The third term of a GP is $4$. Find the product of its first five terms.