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what is the sum of the geometric series?
12,-6,+3,-3/2,.... n=10 so far i have s10=12(1-r10)all over 1-r
i need the ratio and the sum so i can show my work thx
also the sum of the geometric series of 18,-9, 9/2-9/4 ... n=10
multiple choice
A) 575/48
B) 287/24
C) 12
D)37/3
wow thx so much captain helped me a lot now i know how to find the ratio u taught it easier than my teacher thx !!
2 Answers
- 8 years agoFavorite Answer
Let's determine the sum of any geometric series
S = ar + ar^2 + ar^3 + ar^4 + ... + ar^n
Sr = ar^2 + ar^3 + ... + ar^(n + 1)
Sr - S = ar^(n + 1) - ar
S * (r - 1) = ar * (r^n - 1)
S = ar * (1 - r^n) / (1 - r)
So, for our first series we know that ar = 12 and r = -1/2. How do we figure out r? We divide a term by the preceding term. It's that easy.
-6/12 = -1/2
S = 12 * (1 - (-1/2)^10) / (1 - (-1/2))
S = 12 * (1 - 1/1024) / (1 + 1/2)
S = 12 * (1023/1024) / (3/2)
S = 12 * (2/3) * (1023/1024)
S = 4 * 2 * 341 / 1024
S = 341 / 128
S = 18 * (1 - (-1/2)^10) / (1 - (-1/2))
S = 18 * (1 - 1/1024) / (3/2)
S = 18 * (2/3) * (1023/1024)
S = 6 * 2 * 1023 / 1024
S = 3 * 1023 / 256
S = 3069 / 256
- HosamLv 68 years ago
1. a = 12, r = -6 / 12 = -1/2
Sum = S = a (1 - r^10) / (1 - r) = 12 (1 - (-1/2)^10 ) / (1 - (-1/2) ) = 12 (2/3) ( 1 - 1/1024)
= 8 ( 1 - 1/1024) = 8 ( 1023) / 1024 = 1023 / 128
2. For the second series, a = 18, r = -9 / 18 = - 1/2,
S(10) = a (1 - r^10) /(1- r) = (2/3)(18) (1023)/1024 = 12 (1023) / 1024 = 3069 / 256
