The sum of three number in GP is 42. If the first two number are increased by 2 and third is decreased by 4, the resulting form an AP, find the numbers of GP.

Let three numbers be a, b and c

Given $a+b+c=42$                                 …..(i)

Also, $a+2,b+2,c-4\in \mathrm{AP}$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a+2+c-4=2(b+2)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a+c-2=2b+4\end{array}$

$\Rightarrow a+c=2b+6$                         ... (ii) 

From Eq. (i) and Eq. (ii), we ge

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2b+6+b=42\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3b=42-6=36\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b=12\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}ar=12\end{array}$

From Eq. (i), we get 

$\begin{array}{l}a+b+c=42\\ \Rightarrow a+ar+a{r}^2=42\\ \Rightarrow a+12+12r=42\\ \Rightarrow \frac{12}{r}+12+12r=42\\ \frac{1}{r}+1+r=\frac{42}{12}=\frac{7}{2}\\ \Rightarrow 2\left(r^2+r+1\right)=7r\\ \Rightarrow 2r^2-5r+2=0\\ \Rightarrow 2r^2-4r-r+2=0\\ \Rightarrow 2r(r-2)-(r-2)=0\\ \Rightarrow (r-2)(2r-1)=0\\ \Rightarrow r=2,\frac{1}{2}\end{array}$

When $r=2$

         $a=6$

Thus, the sequence is $\{6,12,24\}$

When $r = 1/2$

           $a=24$

Thus, the sequence is $\{24,12,6\}$.