if the sum of n terms of a gp is 255 and nth terms is 128 and common ratio is 2, then first term will be?

if the roots of cubic equation ax cube + bx square + cx + d = 0 are in gp then

option

1. c cube a = b cube d

2. c a cube =b d cube

3. a cube b = c cube d

4. ab cube = cd cube

which one of the following option will b correct

let the first term of GP be a. let the common ratio be r = 2
nth term of GP is 128.
$$\begin{gathered} a.r^{n-1}=128 \\ a.r^n=128.r \\ a.r^n=128*2=256 ............\left(1\right) \end{gathered}$$
sum of n terms of GP is 255
therefore
$$\begin{gathered} \frac{a.\left(r^n-1\right)}{r-1}=255 \\ \frac{a.r^n-a}{2-1}=255 \\ \frac{256-a}{1}=255 \\ a=256-255 \\ a=1 \end{gathered}$$
thus the first term is 1.
$$\begin{gathered} a.r^n=256 \\ 1.2^n=256 \\ {2}^n=2^8 \\ \Rightarrow n=8 \end{gathered}$$ 
thus the total number of terms are 8.

(2)
the given cubic equation is: $a{x}^3+b{x}^2+cx+d=0$
since the roots of the given equation are in GP.
let the roots be $\frac{p}{r}, p , pr$
therefore
product of roots:
$$\begin{gathered} \frac{p}{r}.p.pr=-\frac{d}{a} \\ {p}^3=-\frac{d}{a}..........\left(1\right) \end{gathered}$$
sum of roots:
$$\begin{gathered} \frac{p}{r}+p+pr=-\frac{b}{a} \\ p.\left(\frac{1}{r}+1+r\right)=-\frac{b}{a}............\left(2\right) \\ \frac{p}{r}.p+p.pr+pr.\frac{p}{r}=\frac{c}{a} \\ {p}^2.\left(\frac{1}{r}+r+1\right)=\frac{c}{a}...............\left(3\right) \end{gathered}$$
now dividing the eq(3) by eq(2):
$$\begin{gathered} p=\frac{c/a}{-b/a} \\ p=-\frac{c}{b} \\ {p}^3=-\frac{c^3}{b^3} \\ -\frac{c^3}{b^3}=-\frac{d}{a} [from eq(1) \\ \Rightarrow c^3.a=b^3.d \end{gathered}$$
thus option (a) is correct.

hope this helps you