Question

Find the first term and the common ratio of a geometric progression if the sum of its first three terms is $10.5$ and the difference between the first and the fourth term is equal to $31.5$.

Answer 1

Let the first term be $a$ and common ratio be $r$.

$\therefore a+ar+a{r}^2=10.5\Rightarrow a(1+r+r^2)=10.5-\left(i\right)$

Also,$T_1-T_4=31.5$

$\Rightarrow a-a{r}^3=31.5\Rightarrow a(1-r^3)=31.5\Rightarrow a(1-r)(1+r+r^2)=31.5-\left(ii\right)$

Dividing $\left(ii\right)$ by $\left(i\right)$

$\frac{a(1-r)(1+r+r^2)}{a(1+r+r^2)}=\frac{31.5}{10.5}\Rightarrow 1-r=3\Rightarrow r=-2$

From $\left(i\right)$

$a[1+(-2)+(-2{)}^2]=10.5\Rightarrow a[1-2+4]=10.5\Rightarrow a=\frac{10.5}{3}=3.5$

First term$=3.5$ &common ratio is $-2$.