Find the sum of the geometric series where a1=2, r=-3 and a5=162?

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Find the sum of the geometric series where a1=2, r=-3 and a5=162?

2006-09-28 16:40:43 · 4 answers · asked by airborne 1 in Science & Mathematics ➔ Mathematics

4 answers

Sn = (a1(1 - r^n))/(1 - r)
Sn = (2 * (1 - (-3)^5)))/(1 - (-3))
Sn = (2 * (1 - (-243)))/(1 + 3)
Sn = (2 * (1 + 243))/4
Sn = (2 * 244)/4
Sn = (4 * 122)/4
Sn = 122

ANS : 122

Proof

using a(n) = 2 * (-3)^(n - 1)

a(1) = 2
a(2) = -6
a(3) = 18
a(4) = -54
a(5) = 162

2 - 6 + 18 - 54 + 162 = 122

2006-09-28 17:48:51 · answer #1 · answered by Sherman81 6 · 0⤊ 0⤋

Quick math:

If the geometry sides are 2, -3 and 162, the angle of the 3rd cotangent is 64.

The sum is therefore 623

2006-09-28 23:57:51 · answer #2 · answered by WJ 7 · 0⤊ 0⤋

S5=2[(-3)^5-1]/[(-3)-1]
=122

2006-09-28 23:50:28 · answer #3 · answered by khotl73 2 · 0⤊ 0⤋

the sum is -3

2006-09-28 23:43:03 · answer #4 · answered by Hellfire 2 · 0⤊ 0⤋