what is the sum of the first five ferms of a geometric series with a1=10 and r=1/5? express your answer as an important fraction
in lowest terms without using spaces
1 answer:
The sum of any geometric sequence, (technically any finite set is a sequence, series are infinite) can be expressed as:
s(n)=a(1-r^n)/(1-r), a=initial term, r=common ratio, n=term number
Here you are given a=10 and r=1/5 so your equation is:
s(n)=10(1-(1/5)^n)/(1-1/5) let's simplify this a bit:
s(n)=10(1-(1/5)^n)/(4/5)
s(n)=12.5(1-(1/5)^n) so the sum of the first 5 terms is:
s(5)=12.5(1-(1/5)^5)
s(5)=12.496
as an improper fraction:
(125/10)(3124/3125)
390500/31240
1775/142
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So first of all we can't start by changing them both into improper fractions
So it would be:
5/2 divided by 13/8
From there you would do 5/2 * 8/13 which could convert to
5/1 * 4/13 which would be 20/13
(a-b)(a+b) because when you reverse the sum you do axa and ax-b xx
I believe that x is equal to 6. x=6
Answer: 49°
Step-by-step explanation:
I'm guessing you meant complement? Complementary angles add up to 90°. 90° - 41° = 49°
Answer:
Step-by-step explanation:
Ok