Mathematics - Sequence & Series Question with Solution | TestHub

Given: $3, a, b, c$ are in $A.P$ 

So, the common difference will be $d=a-3$

Now, $b=2a-3$ and $c=3a-6$

Also, $3, a-1, b+1, c+9$ are in $G.P$

So, $3\left(c+9\right)=\left(b+1\right)\left(a-1\right)$

$\Rightarrow 3\left(3a-6+9\right)=\left(2a-3+1\right)\left(a-1\right)$

$\Rightarrow 3\left(3a+3\right)=\left(2a-2\right)\left(a-1\right)$

$\Rightarrow 9a+9=2a^2-4a+2$

$\Rightarrow 2a^2-13a-7=0$

$\Rightarrow a=7, \frac{-1}{2}\left\{\text{rejected}\right\}$

So, $b=2\times 7-3=11 \& c=3\times 7-6=15$

Now, the $A.M$ of $a, b, c$ is given by,

$\Rightarrow A=\frac{a+b+c}{3}$

$\Rightarrow A=\frac{7+11+15}{3}$

$\Rightarrow A=\frac{33}{3}$

$\Rightarrow A=11$