5
10,11,12,13,14 or 8,10,12,14,16
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
x is the first term and d is the difference then x + 3d = 15 and sum of first five terms isx + (x+d) + (x+2d) + (x+3d) + (x+4d)so 5x + 10d = 55 ie x + 2d = 11As x + 3d = 15, d = 4 and x = 3,giving the five terms as 3, 7, 11, 15 and 19
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
10,11,12,13,14 or 8,10,12,14,16