a, a² + 1 and a + 6 are all in arithmetic progression, in which there is a fixed difference d between consecutive terms. This mean we have
a² + 1 = a + d
a + 6 = a² + 1 + d
Eliminate d and solve for a :
(a² + 1) - (a + 6) = (a + d) - (a² + 1 + d)
a² - a - 5 = -a² + a - 1
2a² - 2a - 4 = 0
a² - a - 2 = 0
(a - 2) (a + 1) = 0
a - 2 = 0 or a + 1 = 0
a = 2 or a = -1
