Question 988607
The fourth term of a geometric progression exceeds the third term by 54, and the sum of the second and third term is 36. Find the common ratio if it is positive
<pre>Since {{{a[1] = a[1]}}}, then {{{a[2] = a[1]r}}}, {{{a[3] = a[1]r^2}}}, and {{{a[4] = a[1]r^3}}}

Since {{{a[4]}}} exceeds {{{a[3]}}} by 54, then we can say that: {{{a[1]r^3 - a[1]r^2 = 54}}} 
{{{a[1](r^3 - r^2) = 54}}}
{{{a[1] = 54/(r^3 - r^2)}}} ---------- eq (i)

Since the sum of {{{a[2]}}} and {{{a[3]}}} is 36, then we can say that: {{{a[1]r + a[1]r^2 = 36}}}
{{{a[1](r + r^2) = 36}}}
{{{a[1] = 36/(r + r^2)}}} --------- eq (ii)

Since {{{a[1] = 54/(r^3 - r^2)}}} and {{{a[1] = 36/(r + r^2)}}}, we can then say that: {{{54/(r^3 - r^2) = 36/(r + r^2)}}}
{{{54(r + r^2) = 36(r^3 - r^2)}}} ------ Cross-multiplying
{{{3(r + r^2) = 2(r^3 - r^2)}}} -------- Factoring out GCF, 18
{{{3r + 3r^2 = 2r^3 - 2r^2}}}
{{{2r^3 - 2r^2 - 3r^2 - 3r = 0}}}
{{{2r^3 - 5r^2 - 3r = 0}}}
{{{r(2r^2 - 5r - 3) = r(0)}}}
{{{2r^2 - 5r - 3 = 0}}}
(r - 3)(2r + 1) = 0
Common ratio, or {{{highlight_green(r = 3)}}}		OR          {{{r = - 1/2}}} (ignore, since r MUST be > 0)