arithmetic series (1 Viewer)
- Thread starter rawker
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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.
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.
Last edited:
switchblade87
Member
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.
Edit: Soundly beaten, I'll put the solution in a spoiler.
Sn = n/2 [2a + (n - 1)d]
a = 45 (first term)
d = 2 (difference)
So, you want to find n when the sum = 1365.
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.
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.
Last edited:
switchblade87
Member
Haha, talk it up Riviet!Riviet said:Lol, yeah.