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Common Ratio











































ind the common ratio and write out the first four terms of the geometric sequence {3^n-1/6}
Common ratio is
a1=, a2=, a3=, a4=

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Apr 19, 2011
Geometric Series - Common Ratio
by: Staff


The question:

ind the common ratio and write out the first four terms of the geometric sequence {3^n-1/6}
Common ratio is
a1=, a2=, a3=, a4=

The answer:


A geometric sequence has the (general) form:

a_n = a_1 * (r)^(n - 1)


a_n = a with a subscript of n (this is the nth term in the sequence)

a_1 = a with a subscript of 1 (this is the 1st term in the sequence)

n = number of terms

r = the common ratio


the first four terms in the geometric sequence:


a_n = 3^(n-1)/6


if n = 1, the first term is

a_1 = 3^(1-1)/6

a_1 = 3^(0)/6

a_1 = 1/6


if n = 2, the second term is

a_2 = 3^(2-1)/6

a_2 = 3^(1)/6

a_2 = 3/6

a_2 = 1/2


if n = 3, the third term is

a_3 = 3^(3-1)/6

a_3 = 3^(2)/6

a_3 = 9/6

a_3 = 3/2


if n = 4, the fourth term is

a_4 = 3^(4-1)/6

a_4 = 3^(3)/6

a_4 = 27/6

a_4 = 9/2


a_n = 1/6, 1/2, 3/2, 9/2, . . .

r, the common ratio, can be calculated as follows:
r_n = a_n/a_n-1

(n must be greater than 1)

r_n = r with a subscript of n (this is the common ratio)

a_n = a with a subscript of n (this is the nth term in the sequence)

a_n-1 = a with a subscript of n-1 (this is the n-1 term in the sequence)

using the 2nd and 3rd terms in the sequence to calculate the common ratio

a_2 = 1/2

a_3 = 3/2

r_n = a_n/a_n-1

r_n = a_3/a_2

r_n = (3/2)/(1/2)

r_n = (3/2)*(2/1)

r_n = (3*2)/(2*1)

r_n = 6/2

r_n = 3

r, the common ratio = 3

the final answer is:

the first four terms of the geometric series are:

a_n = 1/6, 1/2, 3/2, 9/2, . . .

the common ration: 3



the standard form of the geometric sequence is:

a_n = a_1 * (r)^(n - 1)

a_n = (1/6) * 3^(n - 1)



Thanks for writing.


Staff
www.solving-math-problems.com


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