Question

If the first $3$ consecutive terms of a geometrical progression are the real roots of the equation $2x^3-19x^2+57x-54=0$, find the sum to infinite number of terms of G.P.

Answer 1

Let the roots be$\frac{a}{r},a,ar$

$\therefore a(1+r+\frac{1}{r})=\frac{19}{2}$

$\therefore a^2(1+r+\frac{1}{r})=\frac{57}{2}$

$\therefore a^3(1+r+\frac{1}{r})=\frac{54}{2}=27$

$\therefore a=3$

$a=(r^2+r+1)\frac{19r}{2}$

$3=(r^2+r+1)\frac{19r}{2}$

$6=(r^2+r+1)19r$

$(2r-3)(3r-2)=0$

$\Rightarrow r=\frac{2}{3}or\frac{3}{2}$

$\therefore$ Numbers $\frac{9}{2},3,2$

$S_{\infty }=\frac{\frac{9}{2}}{1-\frac{2}{3}}=\frac{9}{2}\times \frac{3}{1}=\frac{27}{2}$