Question

find the first four terms of the sequence given by the following.
$a_{n} = 36 - 7(n - 1)$, $n = 1, 2, 3...$

Explanation:

Step1: Find \(a_1\) (n=1)

Substitute \(n = 1\) into \(a_n = 36 - 7(n - 1)\).
\(a_1 = 36 - 7(1 - 1) = 36 - 7\times0 = 36\)

Step2: Find \(a_2\) (n=2)

Substitute \(n = 2\) into the formula.
\(a_2 = 36 - 7(2 - 1) = 36 - 7\times1 = 36 - 7 = 29\)

Step3: Find \(a_3\) (n=3)

Substitute \(n = 3\) into the formula.
\(a_3 = 36 - 7(3 - 1) = 36 - 7\times2 = 36 - 14 = 22\)

Step4: Find \(a_4\) (n=4)

Substitute \(n = 4\) into the formula.
\(a_4 = 36 - 7(4 - 1) = 36 - 7\times3 = 36 - 21 = 15\)

Answer:

36, 29, 22, 15