SOLUTION: Sum of the first 6 terms of a geometric progression equals to 63. Sum of the even terms equals to 42. What's the common ratio and the initial value? Simply wrote, it's s6=63

Algebra ->  Sequences-and-series -> SOLUTION: Sum of the first 6 terms of a geometric progression equals to 63. Sum of the even terms equals to 42. What's the common ratio and the initial value? Simply wrote, it's s6=63      Log On


   

Question 1010896: Sum of the first 6 terms of a geometric progression equals to 63.
Sum of the even terms equals to 42.
What's the common ratio and the initial value?
Simply wrote, it's
s6=63
a2+a4+a6=42
a1=?
r=?
Sum of the odd numbers therefore is A1+A3+A5=21.
I expand a2+a4+a6=42 to a1%2Ar%2Ba1%2Ar%5E3%2Ba1%2Ar%5E5=42 and a1+a3+a5=21 to a1%2Ba1%2Ar%5E2%2Ba1%2Ar%5E4=21.
I can further simplify it to a1%2A%28r%2Br%5E3%2Br%5E5%29=42 and a1%2A%281%2Br%5E2%2Br%5E4%29=21.
I have no idea how to progress further.
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
An=A1 r^(n-1)
A6=A1*r^(n-1);63=A1*r^(5)
A1(r+r^3+r^5)=2A1(1+r^2+r^4)
r^5-2r^4+r^3-2r^2+r-2=0
By synthetic division or graphing, r=2.
Sn=A1(1-r^6)/1-2
63=A1(1-64)/-1
63=63A1
A1=1
A2=2
A3=4
A4=8
A5=16
A6=32
initial value is 1
common ratio is 2