# P1 maths help

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#1
hi i'm having trouble with GP's, i can do the proof but can anyone please show me how to do this with workings, then hopefully i can work out the other questions from that.

"the first term of a geometric series is four times the third term and the common ratio is positive. the sum to infinity is 8"

1. Find the common ratio
2. Calculate, correct to three decimal places the sum of the first tem terms of the series.

Thanks for the help!
0
16 years ago
#2
(Original post by greenie787)
hi i'm having trouble with GP's, i can do the proof but can anyone please show me how to do this with workings, then hopefully i can work out the other questions from that.

"the first term of a geometric series is four times the third term and the common ratio is positive. the sum to infinity is 8"

1. Find the common ratio
2. Calculate, correct to three decimal places the sum of the first tem terms of the series.

Thanks for the help!
1) Looking at the first part:

"the first term of a geometric series is four times the third term and the common ratio is positive. "

A geometric progression takes the form:
a, ar, ar^2, ar^3....
So the first term is a and the third term is ar^2
Therefore, a=4ar^2
So 1=4r^2
Therefore, r^2=1/4
So r=1/2 (as the common ratio is positive, not negative.)

2) We now need to find the value of a:

Sn=a((1-r^n)/(1-r))
When n tends towards infinity:
S=a/(1-r) (As the terms get smaller and the common ratio is positive, it must be less than 1 and larger than 0. Any number less than 1 to a large power is very very small, meaning 1-r^n tends to 1 when n increases towards positive infinity. Substituting in our value for r, we get:
8=a/(1-0.5)
=>a/0.5=8
=>a=4

Now finding S10:
S10=4((1-0.5^10)/(1-0.5)
=7.9921875
0
#3
thank you! i understand it now!
0
16 years ago
#4
(Original post by greenie787)
thank you! i understand it now!
Great, that's good to hear. 0
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