For any sequence of real number $A = {a_{1}, a_{2}, a_{3}, .....},$ a sequence $Δ A$ is defined such that $Δ A = {a_{2} - a_{1}, a_{3} - a_{2}, a_{4} - a_{3}, .....} .$ Suppose that all of the terms of the sequence $Δ (Δ A)$ are 1 and that $a_{19} = a_{92} = 0.$ The units digit in $a_{4}$ is _____.

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Detailed Solution

Suppose that the first term of the sequence $Δ A$ is d then
$Δ A = {d, d + 1, d + 2, ..... (d + (n - 1))}$
Hence $A = (a_{1}, a_{1} + d, a_{1} + d + (d + 1), a_{1} + d + (d + 1) + (d + 2) ......)$
$a_{0} = a_{1} + (n - 1) d + \frac{1}{2} (n - 1) (n - 2)$
So $a_{n}$ is a quadratic polynomial in n.
$S o, a_{n} = \frac{1}{2} (n - 19) (n - 92) S o, a_{3} = \frac{1}{2} (3 - 19) (3 - 92) = 712 a_{4} = \frac{1}{2} (4 - 19) (4 - 92) = \frac{1}{2} (15) (88) = (44) (15) = 660$