How do you find the sum of the arithmetic sequence given 4 + 1 + –2 + –5 + –8 + –11?

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Answer 1 · 2016-07-21T07:25:49

$S_{n} = \frac{n}{2} (11 - 3 n)$ $S_n=n/2(11-3n)$ .

$S_{6} = - 21$ $S_6=-21$ .

Let $a, d & S_{n}$ $a, d & S_n$ denote the $1^{s t}$ $1^(st)$ term, the common difference and the sum of its first $n$ $n$ terms, resp.

Then, we know that,

$S_{n} = \frac{n}{2} {2 a + (n - 1) d}$ $S_n=n/2{2a+(n-1)d}$ .

Here, we have #a=4, d=1-4=-3

Hence, $S_{n} = \frac{n}{2} {8 - 3 (n - 1)} = \frac{n}{2} (8 - 3 n + 3) = \frac{n}{2} (11 - 3 n)$ $S_n=n/2{8-3(n-1)}=n/2(8-3n+3)=n/2(11-3n)$

Referring to our reqd. sum, $n = 6$ $n=6$ , so that,

$S_{6} = \frac{6}{2} (11 - 3 \times 6) = 3 (11 - 18) = 3 (- 7) = - 21$ $S_6=6/2(11-3xx6)=3(11-18)=3(-7)=-21$ .