SOLUTION: Find a10 for the sequence with a1 = 3 and r=2

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Question 1094708: Find a10 for the sequence with a1 = 3 and r=2
Found 2 solutions by ikleyn, jim_thompson5910:
Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.
If it is the geometric progression with the first term 3 and the common ratio of 2,

then the 10-th term is a%5B10%5D = a%5B1%5D%2Ar%5E9 = 3%2A2%5E9 = 1536.


I see many problems AP and GP posted today . . .


On geometric progressions see introductory lessons
    - Geometric progressions
    - The proofs of the formulas for geometric progressions
    - Problems on geometric progressions
    - Word problems on geometric progressions
in this site, written specially for you.

Read them and become an expert in this area.



Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

The nth term of any geometric is
a(n) = a1*r^(n-1)

In our case, a1 = 3 and r = 2 so we have
a(n) = 3*2^(n-1)

Now plug in n = 10
a(n) = 3*2^(n-1)
a(10) = 3*2^(10-1)
a(10) = 3*2^9
a(10) = 3*512
a(10) = 1536

The 10th term is 1536