5
10,11,12,13,14 or 8,10,12,14,16
The first five positive integer terms for 3n + 4 are: 1 = 7 2 = 10 3 = 13 4 = 16 5 = 19
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
5
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.)
It could be any number of your choice. Pick a number and I can find you a quartic (power 4) polynomial which will give the above four and your choice as the first five terms. But the simplest cubic rule is Un = (n3 - 6n2 + 8)/3 for n = 1, 2, 3, ... which gives U5 = 5
4 = -5n + 6n2 if you want to solve for n: 6n2 - 5n - 4 = 0 (3n - 4)(2n + 1) = 0 3n - 4 = 0, n = 4/3 2n + 1 = 0, n = -1/2 n = {-1/2, 4/3}
7
2n(3n + 8)
2,1,0 is th sequence of its terms
6n2+6n+1
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.
6n2
354, 708, 1062, 1416, 1770.