arithmetic series (1 Viewer)

[rawker]

how would i answer this question

How many terms of the series 45+47+49+... give a sum of 1365?

 

[Riviet]

We use the sum of an AP ie Sum of n terms=n/2(2a+(n-1)d), the first time (a) is 45, the common difference (d) is 2, and we are finding the number of terms (n). We also know the total sum so we can sub in the values of a, d and n into the sum of AP formula to find n. I'll start you off:
1365=n/2(90+(n-1)x2)
I find it easy to list out what u know ie a=this, d=that, etc. and then use what you know to find the unknown.

 

[switchblade87]

Well, the formula of the sum of an arithmetic series is:

Sn = n/2 [2a + (n - 1)d]

a = 45 (first term)
d = 2 (difference)
So, you want to find n when the sum = 1365.

Spoiler

So, 1365 = n/2[2 * 45 + (2n - 2)]
2730 = n[90 + 2n - 2]
2730 = n[2n + 88]
0 = 2n² + 88n - 2730
0 = n² + 44n - 1365
0 = (n - 21)(n + 65)
n = 21, -65
Since n can't be negative, n = 21.

Hope that helps.

Edit: Soundly beaten, I'll put the solution in a spoiler.

 

[rawker]

thanks - i wasnt thinking.. i wrote all of that down - maybe i should go to bed lol!

 

[rawker]

thanks for that too switchblade... you guys are too fast haha

 

[Riviet]

Lol, yeah.

 

[switchblade87]

Lol, yeah.

Haha, talk it up Riviet!