Question 1152960



consecutive terms of a geometric progression 

{{{p-3}}},{{{3p-5}}},{{{18-5}}}....not sure about third term, I guess you have {{{18p-5}}}


first express common ratio using given terms:

{{{r=(3p-5)/(p-3)}}}

{{{r=(18p-5)/(3p-5)}}}

{{{r=r}}}=>

{{{(3p-5)/(p-3)=(18p-5)/(3p-5)}}}

{{{(3p-5)(3p-5)=(18p-5)(p-3)}}}

{{{9p^2-30p + 25=18p^2- 59p + 15}}}

{{{18p^2 - 59p + 15-9p^2 + 30p -25=0}}}

{{{9p^2 - 29p - 10=0}}}

use quadratic formula

{{{p=(-(-29)+-sqrt((-29)^2-4*9*(-10))/(2*9))}}}

{{{p=(29+-sqrt(841+360))/18}}}

{{{p=(29+-sqrt(1201))/18}}}


exact solutions:

{{{p=(29+sqrt(1201))/18}}} or
{{{p=(29-sqrt(1201))/18}}}