Find the sum of the first five terms in the geometric series - 3/2 + 1 - 2/3 + ...

a. 55/54
b. -55/54
c. 55/27
d. -55/27

S = ar + ar^2 + ar^3 + ... + ar^n
Sr = ar^2 + ar^3 + ar^4 + ... + ar^n + ar^(n + 1)

Sr - S = ar^(n + 1) - ar
S * (r - 1) = ar * (r^n - 1)
S = ar * (1 - r^n) / (1 - r)

r = -2/3
ar = -3/2
n = 5

S = (-3/2) * (1 - (-2/3)^5) / (1 - (-2/3))
S = (-3/2) * (1 - (-32/243)) / (1 + 2/3)
S = (-3/2) * (1 + 32/243) / (5/3)
S = (-3/2) * (3/5) * (243/243 + 32/243)
S = (-1/2) * (1/5) * 9 * (1/243) * (275)
S = (-1/2) * (1/27) * (55)
S = -55/54

-3/2 + 1 - 2/3 + 4/9 - 8/27 --> Common denominator is 54
-3*(27)/54 + 54/54 - 36/54 + 24/54 - 16/54
= (-81 + 54 - 36 + 24 - 16)/(54)
= (-55)/(54)
= -55/54 --> B

There you go. Hope that helps.

- 3/2 + 1 - 2/3 + ...?

r = 1 by (-3/2) = -2/3

a = -3/2

n = 5

Sum = a (1-r^n)/(1-r)

= (-3/2) (1-(-2/3)^5)/ (1-(-2/3))

= (-3/2) [ 55/81]

= -55/54 [1.0185]

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B