How many terms of the series 3 -6 12 - 24 are n... - JAMB Mathematics 2006 Question
How many terms of the series 3, -6, +12, - 24, + ..... are needed to make a total of 1-28?
A
12
B
10
C
9
D
8
correct option: c
3, -6, +12, -24
a = 3, r = -2
\(8n = \frac{a(1-r^n)}{1-r}\\ ∴1-2^8 = \frac{3(1-(-2^{n-1}))}{1-(-2)}\\ 1-2^8 = \frac{3(1-(-2^{n-1}))}{3}\)
1-28 = 1-(-2)n-1
-28 = -2n-1
8 = n-1
n = 9
a = 3, r = -2
\(8n = \frac{a(1-r^n)}{1-r}\\ ∴1-2^8 = \frac{3(1-(-2^{n-1}))}{1-(-2)}\\ 1-2^8 = \frac{3(1-(-2^{n-1}))}{3}\)
1-28 = 1-(-2)n-1
-28 = -2n-1
8 = n-1
n = 9
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