so 3n2 = 15 ie n2 = 5 so n = sqrt 5
5
10,11,12,13,14 or 8,10,12,14,16
(3n+2)(n+1)
Suppose the first term is a, the second is a+r and the nth is a+(n-1)r. Then the sum of the first five = 5a + 10r = 85 and the sum of the first six = 6a + 15r = 123 Solving these simultaneous equations, a = 3 and r = 7 So the first four terms are: 3, 10, 17 and 24
If that is + 192 then divide all terms by 3 and it is (x+8)(x+8) when factored
All but John Adams served two terms. The total of the first five was nine terms or 36 years (almost - Washington's first term was about an month short.)
7
so 3n2 = 15 ie n2 = 5 so n = sqrt 5
n3 + 3n2 + 4n + 12 = (n3 + 3n2) + (4n + 12) = n2(n + 3) + 4(n + 3) = (n2 + 4)(n + 3).
2,1,0 is th sequence of its terms
5
no clue
What does N equal? Well to solve the problem you would do N+7x1, N+7x2, N+7x 3, N+7x4, N+7x5 to figure out the first five terms.
3
10,11,12,13,14 or 8,10,12,14,16
They are: 7, 10, 13, 16, and 19