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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


Dec 05, 2019
arithmetic sequence NEW
by: Anonymous

A gang has a page on Facebook and 137 Facebook "friends." Suppose that the gang gains 10 more "friends" next week, 13 "friends" the following week, 16 "friends" the week after that, and so on in an arithmetic sequence. If this pattern continues, how many total Facebook "friends" will the gang have 50 weeks from now?

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