SPM Practice Question 1 – 3

Question 1:
The fourth term of a geometric progression is –20. The sum of the fourth and the fifth term is –16. Find the first term and the common ratio of the progression.

Solution:

Question 2:
The fourth and the seventh terms of a geometric progression are 18 and 486 respectively. Find the third term.

Solution:

Question 3:
For a geometric progression, the sum of the first two terms is 30 and the third term exceeds the first term by 15. Find the common ratio and the first term of the geometry progression.

Solution:
$\begin{array}{l}{T}_1+T_2=30\\ a+ar=30\\ a\left(1+r\right)=30---(1)\\ T_3-T_1=15\\ a{r}^2-a=15\\ a\left(r^2-1\right)=15---(2)\\ \\ \frac{\left(2\right)}{\left(1\right)}=\frac{a\left(r^2-1\right)}{a\left(1+r\right)}=\frac{15}{30}\\ \frac{\left(r-1\right)\left(r+1\right)}{1+r}=\frac{1}{2}\to \boxed{\begin{array}{l}\left(r^2-1\right)=\\ \left(r-1\right)\left(r+1\right)\end{array}}\\ r-1=\frac{1}{2}\\ r=\frac{1}{2}+1=\frac{3}{2}\\ \\ \text{From (1),}\\ a\left(1+r\right)=30\\ a\left(1+\frac{3}{2}\right)=30\\ \frac{5a}{2}=30\\ a=12\end{array}$