Hi, the problem im given is to find n when Sn = 366. I know that the sequence is 3+8+13+18+23... and i know the rule for it is An = 5n-2. im pretty sure that if i had the sum of the first and nth term of the sequence i could solve it. could i do ((366/n) *2) = (the sum of the first and last terms) to solve for n? if so, how can i find the sum? any help appresciated
finding n for the given sum Sn(arithmetic series)
- Thread starter asdfg
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n = 1 -- 5*1 - 2
n = 2 -- 5*2 - 2
S2 = 5*(1+2) - 2*2
n = 3 -- 5*3 - 2
S3 = 5*(1+2+3) - 2*3
n = 4 -- 5*4 - 2
S4 = 5*(1+2+3+4) - 2*4
Start to get a picture. We're going to need an expression for the sum of n whole numbers. 1 + 2 + 3 + ... + n = ½*n*(n+1) You should know this. The beauty of this formula is that one of n and n+1 will be even, making the product divisible by 2.
S10 = 5*[½*10*(10+1)] - 2*10
Sn = 5*[½*n*(n+1)] - 2*n
Sometimes, you just have to fool around with it until something makes sense.
So, solve...5*[½*n*(n+1)] - 2*n = 366. I hope you get an integer.
Hint: The answer is NOT -61/5.
n = 2 -- 5*2 - 2
S2 = 5*(1+2) - 2*2
n = 3 -- 5*3 - 2
S3 = 5*(1+2+3) - 2*3
n = 4 -- 5*4 - 2
S4 = 5*(1+2+3+4) - 2*4
Start to get a picture. We're going to need an expression for the sum of n whole numbers. 1 + 2 + 3 + ... + n = ½*n*(n+1) You should know this. The beauty of this formula is that one of n and n+1 will be even, making the product divisible by 2.
S10 = 5*[½*10*(10+1)] - 2*10
Sn = 5*[½*n*(n+1)] - 2*n
Sometimes, you just have to fool around with it until something makes sense.
So, solve...5*[½*n*(n+1)] - 2*n = 366. I hope you get an integer.
Hint: The answer is NOT -61/5.