Question

Find four numbers in G.P. whose sum is 85 and product is 4096.

Answer 1

Let the four numbers in G.P. be$\frac{a}{r^3},\frac{a}{r},ar,a{r}^3$.
So product $=a^4=4096=9\times 512=8\times 8\times 64=8^4\therefore a=8.$
$Sum=8(\frac{1}{r^3}+\frac{1}{r}+r+r^3)=85\Rightarrow 8(r^3+\frac{1}{r^3})+8(r+\frac{1}{r})-85=0....\left(1\right)$
Let $r+\frac{1}{r}=z\therefore {(r+\frac{1}{r})}^3=z^3$
or $r^3+\frac{1}{r^3}+3r\cdot \frac{1}{r}(r+\frac{1}{r})=z^3\therefore r^3+\frac{1}{r^3}=z^3-3z.$
Hence (1) becomes $8(z^3-3z)+8z-85=0$
or $8z^3-16z-85=0...\left(2\right)$
Put $2z=t\therefore t^3-8t=0or(t-5)(t^2+5t+17)=0\therefore t=2z=5or2z=2(r+\frac{1}{r})=5....\left(3\right)$ .The other factor gives imaginary values.
$\therefore$ From (3),$2r^2-5r+2=0(r-2)(2r-1)=0$
$\therefore r=2,\frac{1}{2},$ and $a=8$ Hence
the four numbers are $1,4,16,64or64,16,4,1.$