Mathematics - Sequence & Series Question with Solution | TestHub

Let, the first $3$ terms of the $A.P.$ are $a-d, a, a+d$

$\Rightarrow \mathrm{s}\mathrm{u}\mathrm{m}=3a=33\Rightarrow a=11$

And product $\left(a-d\right) a \left(a+d\right)=a\left(a^2-d^2\right)=1155$

$\Rightarrow 11\left({11}^2-d^2\right)=1155$

$\Rightarrow 121-d^2=\frac{1155}{11}=105$

$\Rightarrow d^2=121-105=16$

$\Rightarrow d=\pm 4$

For $a=11, d=4,$ the $A.P.$ is $7, 11, 15,\dots$

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

So, $T_{11}=7+10\times 4=47$

For $a=11, d=-4,$ the $A.P.$ is $15, 11, 7,\dots$

So, $T_{11}=15+10\left(-4\right)=-25.$