SOLUTION: Please help with this question.how many terms in the series 19+16+13+...must be added to give a sum of 44?

Question 1015443: Please help with this question.how many terms in the series 19+16+13+...must be added to give a sum of 44?
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
the sum of n terms of an arithmetic sequence is,
Sn = (n/2) * (2a1 + (n-1)d) where a1 is the first term to sum, d is the common difference between terms, n is the number of terms to sum
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For our problem, Sn=44, d=-3, a1=19
:
44 = (n/2) * (2(19) + (-3)(n-1))
:
88 = n * ( 38 - 3n + 3)
:
88 = n * (41 - 3n)
:
88 = 41n - 3n^2
:
3n^2 - 41n +88 = 0
:
factor the polynomial
:
(3n-8) * (n-11) = 0
:
there are two possibilities
:
n = 11 or n = 8/3
:
now, we want the nth term, so we can reject n = 8/3
:
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Therefore, 11 terms in the sequence must be summed to get 44
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