SPM Practice Question 15 & 16

Question 15 (2 marks):
It is given that the n^th term of a geometric progression is $T_n=\frac{3r^{n-1}}{2},r\ne k.$  
State
(a) the value of k,
(b) the first term of progression.

Solution:
(a)
k = 0, k = 1 or k = -1 (Any one of these answer).

(b)
$\begin{array}{l}{T}_n=\frac{3}{2}{r}^{n-1}\\ T_1=\frac{3}{2}{r}^{1-1}\\ =\frac{3}{2}{r}^0\\ =\frac{3}{2}\left(1\right)\\ =\frac{3}{2}\end{array}$

Question 16 (4 marks):
It is given that p, 2 and q are the first three terms of a geometric progression.
Express in terms of q
(a) the first term and the common ratio of the progression.
(b) the sum to infinity of the progression.

Solution:
(a)
$\begin{array}{l}{T}_1=p,T_2=2,T_3=q\\ \frac{T_2}{T_1}=\frac{T_3}{T_2}\\ \frac{2}{p}=\frac{q}{2}\\ p=\frac{4}{q}\\ \mathrm{First}\text{ term},T_1=p=\frac{4}{q}\\ \text{Common ratio}=\frac{q}{2}\end{array}$

(b)
$\begin{array}{l}a=\frac{4}{q},r=\frac{q}{2}\\ S_{\infty }=\frac{a}{1-r}\\ =\frac{\frac{4}{q}}{1-\frac{q}{2}}\\ =\frac{4}{q}÷\left[1-\frac{q}{2}\right]\\ =\frac{4}{q}÷\left[\frac{2-q}{2}\right]\\ =\frac{4}{q}\times \frac{2}{2-q}\\ =\frac{8}{2q-q^2}\end{array}$