How to solve this arithmetic series
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How to solve this arithmetic series

[From: ] [author: ] [Date: 12-02-16] [Hit: ]
Please show me the steps on how to do it. I know the solution, just need to learn how to do it .Thanks in advance.For the 1st number in the series, n = 1,......
Sn= -441 for the series 19 + 15 +11 + .... tn. Determine the value of n.

Please show me the steps on how to do it. I know the solution, just need to learn how to do it .

Thanks in advance.

-
A(n) = a + (n-1)d
For the 1st number in the series, n = 1, so
A(1) = a + 0 = 19, so a = 19
Now, d = A(2) - A(1) = 19 - (19 + d) = 4 so d = 4

Series, A(n) = 19 -4n

Sum of arithmetic series:
S(n)= n/2(2a+(n-1)d) put a and d with the above values in this equation
S(n) = (n/2)(38 -4(n-1)) = -441

So, 19n-2n^2 + 2n = -441
2n^2 - 21n -441 = 0
(2n + 21)(n - 21) = 0
n = 21, as n which is the number in the series cannot be negative (-21/2) and cannot be a fraction.
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