SPM Practice Question 2

Question 2:
The third term and the sixth term of a geometric progression are 24 and $7\frac{1}{9}$ respectively. Find
(a) the first term and the common ratio,
(b) the sum of the first five terms,
(c) the sum of the first n terms with n is very big approaching rⁿ ≈ 0.

Solution:
(a)
$\begin{array}{l}\text{Given }{T}_3=24\\ a{r}^2=24\mathrm{...........}\left(1\right)\\ \text{Given }{T}_6=7\frac{1}{9}\\ a{r}^5=\frac{64}{9}\mathrm{...........}\left(2\right)\\ \frac{\left(2\right)}{\left(1\right)}:\frac{a{r}^5}{a{r}^2}=\frac{\frac{64}{9}}{24}\\ r^3=\frac{8}{27}\\ r=\frac{2}{3}\end{array}$

$\begin{array}{l}\text{Substitute }r=\frac{2}{3}\text{ into }\left(1\right)\\ a{\left(\frac{2}{3}\right)}^2=24\\ a\left(\frac{4}{9}\right)=24\\ a=24\times \frac{9}{4}\\ =54\\ \therefore \text{ the first term 54 and the common ratio is }\frac{2}{3}.\end{array}$

(b)
$\begin{array}{l}{S}_5=\frac{54\left[1-{\left(\frac{2}{3}\right)}^5\right]}{1-\frac{2}{3}}\\ =54\times \frac{211}{243}\times \frac{3}{1}\\ =140\frac{2}{3}\\ \\ \therefore \text{ sum of the first five term is }140\frac{2}{3}.\end{array}$

(c)
$\begin{array}{l}\text{When }-1<r<1\text{ and }n\text{ becomes }\\ \text{very big approaching }{r}^n\approx 0,\\ \therefore S_n=\frac{a}{1-r}\\ =\frac{54}{1-\frac{2}{3}}\\ =162\end{array}$
Therefore, sum of the first n terms with n is very big approaching rⁿ ≈ 0 is 162.