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maria17
Group Title
Please help
What is the sum of the arithmetic sequence 134, 122, 110 …, if there are 32 terms?
 11 months ago
 11 months ago
maria17 Group Title
Please help What is the sum of the arithmetic sequence 134, 122, 110 …, if there are 32 terms?
 11 months ago
 11 months ago

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tpaulus Group TitleBest ResponseYou've already chosen the best response.0
Find the general form, then use that to find the last term. then use the sum of an arithmetic sequence equation to find the sum.\[s = n/2 * (n1 + n32)\]
 11 months ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.0
well the sum is simply \[S_{n} = \frac{n}{2}[2a + (n 1)d]\] a = 1st term, n = number of terms and d = common difference in your question you know n = 32, the 1st term is 134 find the common difference. ..d and then substitute and evaluate. the value of d will be negative
 11 months ago

maria17 Group TitleBest ResponseYou've already chosen the best response.0
–1,664 –1,632 –1,600 –1,568
 11 months ago

maria17 Group TitleBest ResponseYou've already chosen the best response.0
Those are the answers.^
 11 months ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.2
S = [134  (0)(12)] + [134  (1)(12)] + [134  (2)(12)] + . . . + [134  (31)(12)] S = (134)(32)  (12)(1 + 2 + 3 + . . . + 31) S = (134)(32)  (12)(31)(32)/2
 11 months ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.2
All good now, @maria17 ?
 11 months ago

maria17 Group TitleBest ResponseYou've already chosen the best response.0
THanksss.
 11 months ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.2
Good luck to you in all of your studies and thx for the recognition! @maria17
 11 months ago

maria17 Group TitleBest ResponseYou've already chosen the best response.0
Can you help me with one more?
 11 months ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.2
Sure, why not? Best though to close this one and just start another post.
 11 months ago
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