**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.

[Note : Using Tn formula ${T}_{n}=a{r}^{n-1}$ to solve for

*a*and

*r*]

*Solution:**T*

_{1}+

*T*

_{2}= 30

*a*+

*ar*

^{ }= 30

*a*(1

*+*

*r*) = 30 —- (1)

*T*

_{3}–

*T*

_{1}= 15

*ar*

^{2}–

*a*= 15

*a*(

*r*

^{2}– 1)

*= 15 —- (2)*

$\begin{array}{l}\frac{\left(2\right)}{\left(1\right)}\text{}\frac{a\left({r}^{2}-1\right)}{a\left(1+r\right)}=\frac{15}{30}\\ \frac{\left(r-1\right)\left(r+1\right)}{1+r}=\frac{1}{2}\to \overline{)\begin{array}{l}\left({r}^{2}-1\right)=\\ \left(r-1\right)\left(r+1\right)\end{array}}\end{array}$

$\begin{array}{l}\text{}r-1=\frac{1}{2}\\ \text{}r=\frac{1}{2}+1=\frac{3}{2}\\ \text{From (1),}a\left(1+r\right)=30\\ \text{}a\left(1+\frac{3}{2}\right)=30\\ \text{}\frac{5a}{2}=30\\ \text{}a=12\end{array}$