Bunuel wrote:

In a sequence, each term starting with the third onward is defined by the following formula \(a_n=a_{n-1}-a_{n-2}\), where n is an integer greater than 2. If the first term of that sequence is 3 and the second term is 4, what is the value of the 70th of that sequence?

A. −4

B. −3

C. −1

D. 3

E. 4

Recursive sequences, where each term is defined in terms of the previous term(s), typically have a useful pattern in test questions I have seen.

Each term has to be found by using previous terms. There is no easy shortcut. Start with what is given, and calculate successive terms until a pattern emerges.

Given: \(A_1 = 3\), \(A_2 = 4\)

\(A_{n} = A_{(n-1)} - A_{(n-2)}\)

\(A_1 = 3\)

\(A_2 = 4\)

\(A_3 = 1\)

\(A_4 = -3\)

\(A_5 = -4\)

\(A_6 = -1\)

\(A_7 = 3\)

(\(A_1\) again)\(A_8 = 4\)

(\(A_2\) again; you can stop) \(A_9 = 1\)

\(A_{10} = -3\)

The easy mistake here would be to decide that \(A_{70} = 3\) because 70 is a multiple of 7, and \(A_7 = 3\). Incorrect.

Instead: find the number of terms in the cycle ("cyclicity"). The pattern repeats every 6 terms. This sequence has a cyclicity of 6.

Divide 70 by 6 to see where \(A_{70}\) "falls" in the cycle.

70/6 has remainder of 4. The R=4 is the subscript number of the term whose value you need. If remainder is 4, Term\(_4\), or fourth term in the cycle of 6, has the same value as Term\(_{70}\).

Or: remainder = ·4· means \(A_{70} = A_{·4·}\)

\(A_{70} = A_4 = -3\)

Answer B

*

\(A_3\) = \((A_2) - (A_1)\)

\(A_3\) = \((4) - (3) = 1\)

\(A_3 = 1\)
_________________

In the depths of winter, I finally learned

that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"