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Mathematics 16 Online
OpenStudy (anonymous):

8. What is the sum of the arithmetic sequence 140, 136, 132..., if there are 35 terms?

jimthompson5910 (jim_thompson5910):

Can you tell me what the last term is?

OpenStudy (anonymous):

They dont say. I have to figure it out I guess.

jimthompson5910 (jim_thompson5910):

Do you know how to figure out the last term?

OpenStudy (anonymous):

No clue:/

jimthompson5910 (jim_thompson5910):

the general term is an = a1 + d(n-1) The first term is 140, the common difference is -4, so a1 = 140 and d = -4

jimthompson5910 (jim_thompson5910):

therefore we can say an = 140 + (-4)(n-1) Now we want the 35th term of this sequence, so plug in n = 35 an = 140 + (-4)(35-1) an = 140 + (-4)(34) an = 140 - 136 an = 4

jimthompson5910 (jim_thompson5910):

so the last term (the 35th term) is 4

jimthompson5910 (jim_thompson5910):

so.. Sum of first 35 terms S = (n*(a1 + an))/2 S = (35*(140 + 4))/2 S = ???

OpenStudy (anonymous):

2,520!?

jimthompson5910 (jim_thompson5910):

yep, you nailed it

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