d = 6
233 - 53 = 180 ... so, 31 terms
s = (233 + 53) * 31 / 2
Find the sum of each series by using appropriate formulas. Show your work.
MY WORK:
(N is the last number in the sequence)
1+2+3+...+230
n(n+1)/(1*2)
4+8+12+...+360
n(n+4)/)4*2)
1+3+5+...+123
((n+1)/2))^2
The problem:
53+59+65+...+233
n(n+6)/(6*2)
233(239)/12 = decimal???
There can't be a decimal when finding a sum of a sequence. What did I do wrong?
2 answers
I have no idea what that first part of your post is supposed to be.
your problem:
53+59+65+...+233
We have to know how many terms.
a=53, d=6, n= ?
term(n) = a + (n-1)d
233 = 53 + 6(n-1)
180 = 6n - 6
6n = 186
n = 31
sum(31) = (31/2)(first + last)
= (31/2)(53+233) = 4433
which is what Scott's answer would give you
your problem:
53+59+65+...+233
We have to know how many terms.
a=53, d=6, n= ?
term(n) = a + (n-1)d
233 = 53 + 6(n-1)
180 = 6n - 6
6n = 186
n = 31
sum(31) = (31/2)(first + last)
= (31/2)(53+233) = 4433
which is what Scott's answer would give you
1 answer