10,11,12,13,14 or 8,10,12,14,16
The sequence 4n + 7 represents a linear sequence where n is the position in the sequence. To find the first five terms, substitute n with 1, 2, 3, 4, and 5 respectively. Thus, the first five terms are 11, 15, 19, 23, and 27.
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
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
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
10,11,12,13,14 or 8,10,12,14,16