Mathematics - Sequence & Series Question with Solution | TestHub

We know that, the $n^{th}$ term of a $G.P.$ with first term $a$ and common ratio $r$ is $t_n=a{r}^{n-1}.$

Then, $b=ar$ and $c=a{r}^2$

Now, $3a, 7b$ and $15c$ are in $A.P.$

$\Rightarrow 14b=3a+15c$

$\Rightarrow 14\left(ar\right)=3a+15\left(a{r}^2\right)$

$\Rightarrow 14r=3+15r^2$

$\Rightarrow 15r^2-14r+3=0$

$\Rightarrow \left(3r-1\right)\left(5r-3\right)=0$

$\Rightarrow r=\frac{1}{3}$ or $r=\frac{3}{5}$

The only acceptable value is $r=\frac{1}{3},$ because $r\in \left(0, \frac{1}{2}\right]$

$\therefore$ the first term of the $A.P.$ is $A=3a$ and the common difference is $d=7b-3a=7ar-3a=\frac{7}{3}a-3a=-\frac{2}{3}a$

Also, we know that the $n^{th}$ term of an $A.P.$ with first term $A$ and common difference $d$ is $T_n=A+\left(n-1\right)d$

$\therefore T_4=3a+\left(4-1\right)\left(-\frac{2}{3}a\right)= 3a-2a=a.$