SOLUTION: Find all geometric sequences in which t3 + t4 = 36 and t5 + t6 = 144 I know one could be t1=3 t2=6 t3=12 t4=24 t5=48 t6=96 That would fit the question, but i'm assuming the

Algebra ->  Sequences-and-series -> SOLUTION: Find all geometric sequences in which t3 + t4 = 36 and t5 + t6 = 144 I know one could be t1=3 t2=6 t3=12 t4=24 t5=48 t6=96 That would fit the question, but i'm assuming the      Log On


   

Question 762771: Find all geometric sequences in which t3 + t4 = 36 and t5 + t6 = 144

I know one could be t1=3 t2=6 t3=12 t4=24 t5=48 t6=96
That would fit the question, but i'm assuming there's more than one geometric sequence that fits? There is two questions like this on my assignment. Don't know how to go about finding multiple sequences
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
The way to solved it could be this one:

Ar%5E2+%2B+Ar%5E3+=+36 and
Ar%5E4+%2B+Ar%5E5+=+144

These can be factored to
Ar%5E2+%28r%2B1%29+=+36 and
Ar%5E4+%28r%2B1%29+=144
Therefore,
r%2B1+=+36%2F%28Ar%5E2%29 and
r%2B1+=+144%28Ar%5E4%29

Setting these equal to each other and cross multiplying gives
144Ar%5E2+=+36Ar%5E4, or

4Ar%5E2=Ar%5E4

One solution is for r to be 0, but that obviously won't work.

So, dividing through by Ar%5E2 gives 4=r%5E2, so r can be either 2 or -2.

Using these with the previous equations gives the value of A,
and therefore the sequence.
A%2A2%5E2+%2B+A%2A2%5E3+=+36 ..=>4A+%2B+8A+=+36..=>12A+=+36..=>A+=+36%2F12..=>A+=+3
same here
A%2A2%5E4+%2B+A%2A2%5E5+=+144..=>16A%2B32A=144..=>48A=144..=>A=144%2F48..=>A=3
geometric sequences is
Ar%5E0=t1=3%2A1=3+
Ar%5E1=t2=3%2A2=6+
Ar%5E2=t3=+3%2A4=12
Ar%5E3=t4=3%2A8=24
Ar%5E4=t5=3%2A16=48
Ar%5E5=t6=3%2A32=96
Ar%5E6=t7=3%2A64=192...and so on
check if t3+%2B+t4+=+36 and t5+%2B+t6+=+144
12+%2B+24+=+36
36=+36
48+%2B+96+=+144
144+=+144