Question 762771
The way to solved it could be this one:
 
{{{Ar^2 + Ar^3 = 36}}} and 
{{{Ar^4 + Ar^5 = 144}}}
 
These can be factored to 

{{{Ar^2 (r+1) = 36}}} and 
{{{Ar^4 (r+1) =144}}}

Therefore, 

{{{r+1 = 36/(Ar^2)}}} and 
{{{r+1 = 144(Ar^4)}}}
 
Setting these equal to each other and cross multiplying gives 

{{{144Ar^2 = 36Ar^4}}}, or
 
{{{4Ar^2=Ar^4}}}
 
One solution is for {{{r}}} to be {{{0}}}, but that obviously won't work.
 
So, dividing through by {{{Ar^2}}} gives {{{4=r^2}}}, so {{{r}}} can be either {{{2}}} or {{{-2}}}.
 
Using these with the previous equations gives the value of {{{A}}}, 
and therefore the sequence.

{{{A*2^2 + A*2^3 = 36}}} ..=>{{{4A + 8A = 36}}}..=>{{{12A = 36}}}..=>{{{A = 36/12}}}..=>{{{A = 3}}}

same here

{{{A*2^4 + A*2^5 = 144}}}..=>{{{16A+32A=144}}}..=>{{{48A=144}}}..=>{{{A=144/48}}}..=>{{{A=3}}}

geometric sequences is

{{{Ar^0=t1=3*1=3 }}}
{{{Ar^1=t2=3*2=6 }}}
{{{Ar^2=t3= 3*4=12}}} 
{{{Ar^3=t4=3*8=24}}}
{{{Ar^4=t5=3*16=48}}}
{{{Ar^5=t6=3*32=96}}}
{{{Ar^6=t7=3*64=192}}}...and so on

check if  {{{t3 + t4 = 36}}} and {{{t5 + t6 = 144}}}

{{{12 + 24 = 36}}}
{{{36= 36}}}

{{{48 + 96 = 144}}}
{{{144 = 144}}}