an arithmetic sequence of positive integers has a common difference that is three times the first term. If the sum of the first five terms is equal to the absolute difference between the first term and its square, what is the first term in the sequence
Let the first term = F
Common difference=3F
F^2 - F = F + 3F*4
F^2 = 2F + 12F
F^2 = 14F, solve for F
F = 14 - First term
Sorry!. I missed one F:
Let the first term = F
Common difference=3F
F^2 - F = 2F + 3F*4
F^2 = 3F + 12F
F^2 = 15F, solve for F
F = 15 - First term
Let the first term = a
Sum of first 5 terms =
a + (a + 3a) + (a + 2*3a) + (a + 3*3a) + ( a + 4*3a) = 35a
So
l a - a^2 l = 35a
Then either......
a - a^2 = 35a or a^2 - a = 35a
a^2 + 34a = 0 a^2 - 36a = 0
a(a + 34) =0 a ( a - 36) = 0
a = 0 a = -34 or a = 36
Since a is positive......then a = 36