How many terms of the series -3 -1 + 1 +..... add up to 165?
Question 1 Report
How many terms of the series -3 -1 + 1 +..... add up to 165?
Answer Details
The series \(-3,\ -1,\ 1,\ \dots\) is arithmetic with first term \(a=-3\) and common difference \(d=2\).
\[S_n = \frac{n}{2}\big[2a + (n-1)d\big] = \frac{n}{2}\big[-6 + 2(n-1)\big] = \frac{n}{2}(2n-8) = n(n-4)\]Set \(S_n = 165\):
\[n(n-4) = 165 \;\Rightarrow\; n^2 - 4n - 165 = 0\] \[n = \frac{4 \pm \sqrt{16 + 660}}{2} = \frac{4 \pm \sqrt{676}}{2} = \frac{4 \pm 26}{2}\]Taking the positive value, \(n = \dfrac{30}{2} = 15\).
15 terms are required.
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