Maths Solve the Following Question.(2 Marks)

The terms 0.4, 0.04, 0.004,… are in G.P.

$\therefore a=0.4, r=\frac{0.07}{0.7}=0.1,|r|=10.11<1$

∴ Sum to infinity exists.

∴ Sum to infinity

$\begin{aligned}=2+\frac{a}{1-r} & =2+\frac{0.4}{1-0.1} \\ & =2+\frac{0.4}{0.9} \\ & =2+\frac{4}{9}=\frac{22}{9}\end{aligned}$

The terms 0.7, 0.07, 0.007,… are in G.P.

$\therefore a=0.7, r=\frac{0.07}{0.7}=0.1,|r|=|0.1|<1$

∴ Sum to infinity exists.

∴ Sum to infinity

$\begin{aligned}=\frac{a}{1-r} & =\frac{0.7}{1-0.1} \\ & =\frac{0.7}{0.9} \\ & =\frac{7}{9}\end{aligned}$

Let $\frac{q}{p}>1$

$\begin{aligned} & S_n=\frac{a\left(r^n-1\right)}{r-1}, \text { for } r>1 \\ \therefore \quad & S_n=\frac{p\left[\left(\frac{q}{p}\right)^n-1\right]}{\frac{q}{p}-1}=\frac{p^2}{q-p}\left[\left(\frac{q}{p}\right)^n-1\right]\end{aligned}$

$\therefore \quad \frac{t_2}{t_1}=\frac{t_3}{t_2}$

$\therefore \quad \frac{x}{\frac{4}{3}}=\frac{\frac{4}{27}}{x}$

$\begin{array}{ll}\therefore & x^2=\frac{4}{3} \times \frac{4}{27} \\ \therefore & x^2=\frac{16}{81} \\ \therefore & x= \pm \frac{4}{9}\end{array}$

$\begin{array}{ll} & t_n=a r^{n-1} \\ \therefore \quad & 5^{10}=5 \times 5^{(n-1)} \\ \therefore \quad & 5^{10}=5^{(1+n-1)}\end{array}$

$\begin{array}{ll}\therefore & 5^{10}=5^n \\ \therefore & n=10\end{array}$

$\therefore \quad n=10$

$\therefore \quad 5^{10}$ is the $10^{\text {th }}$ term of the G.P.

Alternate Method:

$\begin{aligned} & t_1=5, t_2=25=5^2, t_3=125=5^3, t_4=625=5^4, \\ \therefore \quad & t_{10}=5^{10}\end{aligned}$