Question

Find the sum of the series shown below.

5 + 9 + 13 + ... + 161

Answer 1

@cinar please explain your answer, as per the new code of conduct on OS
just posting answers is not acceptable in situations like these. We strive to teach here, not show off.

Answer 2

@devin.cobbs there is a formula for this, that I always forget, but can be found on Wikipedia
http://en.wikipedia.org/wiki/Arithmetic_progression

Answer 3

ok thanks alot!

Answer 4

If you know the n then n/2 * (First term +last term) is sufficient for this problem.

Answer 5

I believed the formula you want is\[S=\frac n2[2a_1+(n-1)d]\]where \(d\) is the common difference between each term, \(n\) is the number of terms, and \(a_1\) is the first term in the series.
now just plug in and solve, I think...

-oh thanks FFM
@devin.cobbs listen to FFM, he's better at this kind of thing than me.
I always just work series from scratch from the sigma sign...

Answer 6

@devin.cobbs Do you know, how to find n ?

Answer 7

i see it, the formula @turingtest has it above.

Answer 8

\[\S=\frac{(last\quad term+first \quad term)* (last\quad term-first \quad term +common\quad difference) }{2*common\quad difference}\]

Answer 9

\[\small\S=\frac{(last\quad term+first \quad term)* (last\quad term-first \quad term +common\quad difference) }{2*common\quad difference}\]

Answer 10

for arithmetic sum, always use this formula..

Answer 11

so for your question..
\[\S=\frac{(161+5)*(161-5+4)}{2*4}=3320\]