Question 970124
If in a geometric progression,the second term exceeds the first term by 20 and the fourth term exceeds the second term by 15,find the possible values of the first term.
Solution:
Let x = first term of a geometric progression and r = common ratio.
The terms of a geometric progression in terms of x and r are
x, xr, {{{xr^2}}}, {{{xr^3}}}, ...
Therefore,
xr - x = 20  and  {{{xr^3}}} - xr = 15
{{{x = 20/(r-1)}}}  and {{{x = 15/(r^3-r)}}}
{{{20/(r-1) = 15/(r^3-r)}}}
{{{20(r^3-r) = 15(r-1)}}}
{{{20r^3-35r+15=0}}}
5(r-1)(2r-1)(2r+3) = 0
r = 1, 1/2, -3/2
If r = 1, then x is undefined
If r = 1/2, then x= {{{20/(1/2-1) = -40}}}
If r = -3/2, then x = {{{20/(-3/2-1) = -8}}}
Answer: -40 or -8