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ARITHMETIC SEQUENCE WORD PROBLEM HELP!!











































IsIf the 100th term of an arithmetic sequence is 590, and its common difference is 6, then
its first term a1= ,
its second term a2= ,
its third term a3= .

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Apr 18, 2011
Arithmetic Sequence – Solving a Word Problem
by: Staff


The question:

If the 100th term of an arithmetic sequence is 590, and its common difference is 6, then
its first term a1= ,
its second term a2= ,
its third term a3= .

The answer:

If the 100th term of an arithmetic sequence is 590, and its common difference is 6, then
its first term a1= ,
its second term a2= ,
its third term a3= .


An arithmetic sequence has the form:

a_n = a_1 + (n - 1) * (d)


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

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

n = number of terms

d = difference between consecutive terms


Applying the notation to your problem

a_100 = 590

d = difference between consecutive terms = 6


a_n = a_1 + (n - 1) * (d)

a_100 = a_1 + (100 - 1) * (6)

a_1 + (100 - 1) * (6) = 590

a_1 + (99) * (6) = 590

a_1 + 594 = 590

a_1 + 594 – 594 = 590 – 594

a_1 + 0 = 590 – 594

a_1 = 590 – 594

a_1 = -4

the formula for the sequence is:

a_n = a_1 + (n - 1) * (d)

a_n = -4 + (n - 1) * (6)


the first term of the sequence is (n = 1):

a_1 = -4



the second term of the sequence is (n = 2):

a_2 = -4 + (2 - 1) * (6)

a_2 = -4 + 6

a_2 = 2


the third term of the sequence is (n = 3):

a_3 = -4 + (3 - 1) * (6)

a_3 = -4 + 2 * (6)

a_3 = -4 + 12

a_3 = 8



Thanks for writing.


Staff
www.solving-math-problems.com


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