NIMCET 2017 Question Paper with solutions

Consider, three positive number $\mathit{a}-\mathit{d},\mathit{a},\mathit{a}+\mathit{d}$ are in $\mathit{A}\mathit{P}$
Three positive number whose sum is 21 .
$\Rightarrow \mathit{a}-\mathit{d}+\mathit{a}+\mathit{a}+\mathit{d}=21$
$\Rightarrow \mathit{a}=7$
Hence, three positive number becomes $7-\mathit{d},7,7+\mathit{d}$.
It is given, If $2,2,14$ are added to them respectively then resulting number are in geometric progression.
$\Rightarrow 9-\mathit{d},9,21+\mathit{d}$ are in GP.
$\Rightarrow \frac{9}{9-\mathit{d}}=\frac{21+\mathit{d}}{9}$
$\Rightarrow 81=\left(21+\mathit{d}\right)\left(9-\mathit{d}\right)$
$\Rightarrow {\mathit{d}}^2+12\mathit{d}-108=0$
$\Rightarrow \left(\mathit{d}+18\right)\left(\mathit{d}-6\right)=0$
$\Rightarrow \mathit{d}=-18$ and $\mathit{d}=6$
When $\mathit{d}=-18$, the three numbers are $25,7,-11$
When $\mathit{d}=6$, the three numbers are $1,7,13$