What is the common ratio of the geometric sequence: 128, -16, 2, -1/4, …?
What is the common ratio of the geometric sequence: 128, -16, 2, -1/4, …?
Option A: -8
Option B: 8
Option C: -1/8
Option D: 1/8
Update:Write the series in expanded form and find the sum.
5
∑(-6k+14)
k=1
(6 ∙ 1 14) + (6 ∙ 5 + 14) = 8
(6 ∙ 1 14) (6 ∙ 5 + 14) = 24
(6 ∙ 1 14) (6 ∙ 2 14) (6 ∙ 3 14) (6 ∙ 4 14) (6 ∙ 5 14) = 36
(6 ∙ 1 14) + (6 ∙ 2 14) + (6 ∙ 3 14) + (6 ∙ 4 14) + (6 ∙ 5 14) = 20
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Verified answer
common ratio = -16/128 = -1/8 which is option C
by substituting values of k ,we get this series:
8 , 2 , -4 , -10 , -16
now this is an arithmetic sequence with common difference(d) = -6
sum = n/2(first term+last term)
sum= 5/2(8 -16)
sum=-20 which is NONE of the given options
Hope This Helps :D
1- option c is correct
common ratio of the geometric sequence is -1/8
because common ratio r =second term / first term = -16/128=-1/8
2- For your second question
if you expand this series
8 , 2 , -4 , -10 , 16
hence in this series common difference are same
so this series is an AP
apply sum formula of an AP
Sn=n/2 [ a + an] where an is the last term
Sn=5/2 [8 +(-16)]
Sn=5/2 * -8
simplify
Sn = -20
hence sum of this series is -20
128 = 2^7
16 = 2^4
2 = 2^1
1/4 = 2^(-2)
So the formula is (-1)^(n+1) * 2^(10-3n) for n >= 1
alternatively, 128 * (-1/8)^(n-1) = 2^7 * ((-2)^(-3))^(n-1)
the n+1th term is always -1/8 of the nth term. I think this is the common ratio you are looking for.