Question

The sum of the first. two terms of a geometric, progression is $15$. The first term exceeds the common ratio of the progression by $\frac{25}{3}$. Find the fourth term of the progression.

Answer 1

Let the first term be $a_1$ & common ratio be $r$.

$\therefore T_1+T_2=15\Rightarrow a_1+a_{1}r=15\Rightarrow a_1(1+r)=15-\left(i\right)$

Also, $a_1-r=\frac{25}{3}\Rightarrow a_1=\frac{25}{3}+r$

Replacing $a_1=\frac{25}{3}+r$ in $\left(i\right)$

$\Rightarrow (\frac{25}{3}+r)(1+r)=15\Rightarrow \frac{25}{3}+r+\frac{25}{3}r+r^2=15\Rightarrow r^2+\frac{25}{3}r+r+\frac{25}{3}-15=0\Rightarrow r^2+\frac{28}{3}r+\frac{25-45}{3}=0\Rightarrow {3r}^2+28r-20=0$

Using Quadratic equation:

$a=3,b=28,=-20\therefore r=\frac{-28\pm \sqrt{{\left(28\right)}^2-4\times 3\times (-20)}}{2\times 3}=\frac{-28\pm \sqrt{784+240}}{6}=\frac{-28\pm \sqrt{1024}}{6}=\frac{-28-32}{6},\frac{-28+32}{6}=\frac{-60}{6},\frac{4}{6}=-10,\frac{2}{3}$

Integrating negative term,

$r=\frac{2}{3}{a}_1=(1+\frac{2}{3})=15\Rightarrow a_1\left(\frac{3+2}{3}\right)=15\Rightarrow \frac{5a_1}{3}=15\Rightarrow a_1=\frac{15\times 3}{5}=9\therefore T_4=a_1{r}^{4-1}=a_1{r}^3=9\times {\left(\frac{2}{3}\right)}^2=\frac{8}{3}=2\frac{2}{3}$