the fourth term of a GP is greater than the first term which is positive by 372 the third term is greater than second by 60. calculate the common ratio and the first term of the progression

Dear Student,


Let the 1st term od the GP=a
And the common ratio=r
So the terms of the GP will be $a, ar, a{r}^2, a{r}^3$
Given that the difference between 4th and the 1st term will be 372.
That means 
$$\begin{gathered} a{r}^3-a=372 \\ \Rightarrow a(r^3-1)=372 \\ \Rightarrow a(r-1)(r^2+r+1)=372.....\left(1\right) \end{gathered}$$
Again it is given that difference between 3rd and the 2nd term will be 60.
That means
$$\begin{gathered} a{r}^2-ar=60 \\ \Rightarrow ar(r-1)=60 \\ \Rightarrow a(r-1)=\frac{60}{r} \end{gathered}$$
Now put the value of a(r-1) in equation 1
$$\begin{gathered} \frac{60}{r}(r^2+r+1)=372 \\ \Rightarrow 60(r^2+r+1)=372r \\ \Rightarrow 60r^2+60r+60-372r=0 \\ \Rightarrow 60r^2-312r+60=0 \\ Deviding 12 on both the sides \\ \Rightarrow 5r^2-26r+5=0 \end{gathered}$$
After solving this quadratic equation we will get r=5.
That means common ratio is 5.
We earlier got, $$\begin{gathered} a(r-1)=\frac{60}{r} \\ \Rightarrow a(5-1)=\frac{60}{5} \\ \Rightarrow 4a=12 \\ \Rightarrow a=3 \end{gathered}$$
So here the 1st term of GP=3.
So the progression will be 3,15,75,375. (Answer)

Regards,