Question

If the sum of the first n terms of the series,4+7+10+... is 209, find n

Answer 1

Looks like an AS, with a = 4 and d = 3
(n/2)(2a + (n-1)d) = sum(n)
(n/2)(8 + 3(n-1)) = 209
n(8 + 3n - 3) = 418
5n + 3n^2 - 418 = 0
3n^2 + 5n - 418 = 0

solve for n, reject the negative answer

Answer 2

difference = 3 = d
first term = 4 = a
https://www.mathsisfun.com/algebra/sequences-sums-arithmetic.html
209 = (n/2) [ 4 + (n-1)3 ]
209 = (n/2) [ 1+3n ]
418 = n + 3 n^2
3 n^2 + 1 n - 418 = 0
about 11.64
hmmm, typo or my mistake ?

Answer 3

209 = (n/2) [ 4 + (n-1)3 ]
4 should be 8, 2 n not n
209 = (n/2) [ 8 + (n-1)3 ]
418 = 5n + 3 n^2
3 n^2 + 5 n - 418 = 0

Answer 4

Yh true