Asked by Fading
Describe an infinite geometric series with a beginning value of 2 that converges to 10. What are the first 4 terms of the series?
Last question and my mind is fried, can someone please help me? Thanks
Last question and my mind is fried, can someone please help me? Thanks
Answers
Answered by
Bosnian
S ∞ =
∞
∑ = a * r ^ n = a ⁄ ( 1 – r )
n=1
In this case a = 2 , infinite sum = 10 so:
S ∞ = 10 = a ⁄ ( 1 – r )
10 = 2 ⁄ ( 1 – r ) Multiply both sides by 1 - r
10 * ( 1 – r ) = 2
10 * 1 – 10 * r = 2
10 - 10 r = 2 Subtract 10 to both sides
10 - 10 r - 10 = 2 - 10
- 10 r = - 8 Divide both sides by - 10
- 10 r / - 10 = - 8 / - 10
r = 0.8
The n-th term of a geometric series with initial value a and common ratio r is given by:
an = a * r ^ ( n − 1 )
a1 = a * r ^ ( n − 1 ) = 2 * 0.8 ^ ( 1 − 1 ) = 2 * 0.8 ^ 0 = 2 * 1 = 2
a2 = a * r ^ ( n − 1 ) = 2 * 0.8 ^ ( 2 − 1 ) = 2 * 0.8 = 1.6
a3 = a * r ^ ( n − 1 ) = 2 * 0.8 ^ ( 3 − 1 ) = 2 * 0.8 ^ 2 = 2 * 0.64 = 1.28
a4 = a * r ^ ( n − 1 ) = 2 * 0.8 ^ ( 4 − 1 ) = 2 * 0.8 ^ 3 = 2 * 0.512 = 1.024
∞
∑ = a * r ^ n = a ⁄ ( 1 – r )
n=1
In this case a = 2 , infinite sum = 10 so:
S ∞ = 10 = a ⁄ ( 1 – r )
10 = 2 ⁄ ( 1 – r ) Multiply both sides by 1 - r
10 * ( 1 – r ) = 2
10 * 1 – 10 * r = 2
10 - 10 r = 2 Subtract 10 to both sides
10 - 10 r - 10 = 2 - 10
- 10 r = - 8 Divide both sides by - 10
- 10 r / - 10 = - 8 / - 10
r = 0.8
The n-th term of a geometric series with initial value a and common ratio r is given by:
an = a * r ^ ( n − 1 )
a1 = a * r ^ ( n − 1 ) = 2 * 0.8 ^ ( 1 − 1 ) = 2 * 0.8 ^ 0 = 2 * 1 = 2
a2 = a * r ^ ( n − 1 ) = 2 * 0.8 ^ ( 2 − 1 ) = 2 * 0.8 = 1.6
a3 = a * r ^ ( n − 1 ) = 2 * 0.8 ^ ( 3 − 1 ) = 2 * 0.8 ^ 2 = 2 * 0.64 = 1.28
a4 = a * r ^ ( n − 1 ) = 2 * 0.8 ^ ( 4 − 1 ) = 2 * 0.8 ^ 3 = 2 * 0.512 = 1.024
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