(i) How many terms of the G.P. 3, 3 $$^{2}, 3^{3}, \ldots$$ are needed to give the sum 120?
From the question, it is given that,
Terms of the G.P. $$3,3^{2}, 3^{3}, \ldots$$
Sum of the terms $$=120$$
The first term $$a=3$$
$$\begin{aligned}r &=3^{2} / 3 \\&=9 / 3 \\&=3\end{aligned}\\$$
We know that, $$S_{n}=a\left(r^{n}-1\right) / r-1=120$$
$$\begin{array}{l}3\left(3^{n}-1\right) /(3-1)=120 \\3\left(3^{n}-1\right) / 2=120\end{array}\\$$
By cross multiplication we get,
$$\begin{array}{l}3^{n}-1=(120 \times 2) / 3 \\3^{n}-1=240 / 3 \\3^{n}-1=80 \\3^{n}=80+1 \\3^{n}=81 \\3^{n}=3^{4}\end{array}\\$$
Therefore, $$n=4$$
(ii) How many terms of the G.P. $$1,4,16, \ldots$$ must be taken to have their sum equal to $$341 ?$$
From the question, it is given that,
Terms of the G.P. $$1,4,16, \ldots$$
Sum of the terms $$=341$$
The first term $$a=1$$
$$\begin{aligned}r &=4 / 1 \\&=4\end{aligned}\\$$
We know that,
$$S_{n}=a\left(r^{n}-1\right) / r-1=341$$
$$\begin{array}{l}1\left(4^{n}-1\right) /(4-1)=341 \\1\left(4^{n}-1\right) / 3=341\end{array}\\$$
By cross multiplication we get,
$$\begin{array}{l}4^{n}-1=(341 \times 3) \\4^{n}-1=1023 \\4^{n}=1023+1 \\4^{n}=1024\end{array}\\$$
$$4^{n}=4^{5}\\$$
Therefore, $$n=5$$