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so 3n2 = 15 ie n2 = 5 so n = sqrt 5
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.
(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