Step 1
Consider the given series: \(67, 63, 59, 53...\)
Step 2
The terms of the series can be denoted as follows:
\(a_1 = 67,\)

\(a_2 = 63,\)

\(a_3 = 59,\)

\(a_4 = 55,\) Step 3 Now check the type of the series: \(a_1 - a_2 = 67 - 63 = 4,\)

\(a_2 - a_3 = 63 - 59 = 4,\)

\(a_3 - a_4 = 59 - 55 = 4,\) Since \(a_1 - a_2 = a_2 - a_3 = a_3 - a_4 = 4,\) Therefore series is arithmetic Step 4 The formula for the nth term is determined as follows: \(a_1 = 67,\)

\(a_2 = 63,\)

\(a_2 = a_1 - 4 \cdot (2-1),\)

\(a_3 = 59,\)

\(a_3 = a_1 -4(3-1),\) Now determine thr \(6^(th)\) term sa follows: \(a^{60} = a_1 - 4(60 - 1)\)

\(a^{60} = 67 - 4 \cdot 59\)

\(d^{60} = 169\)

\(a_2 = 63,\)

\(a_3 = 59,\)

\(a_4 = 55,\) Step 3 Now check the type of the series: \(a_1 - a_2 = 67 - 63 = 4,\)

\(a_2 - a_3 = 63 - 59 = 4,\)

\(a_3 - a_4 = 59 - 55 = 4,\) Since \(a_1 - a_2 = a_2 - a_3 = a_3 - a_4 = 4,\) Therefore series is arithmetic Step 4 The formula for the nth term is determined as follows: \(a_1 = 67,\)

\(a_2 = 63,\)

\(a_2 = a_1 - 4 \cdot (2-1),\)

\(a_3 = 59,\)

\(a_3 = a_1 -4(3-1),\) Now determine thr \(6^(th)\) term sa follows: \(a^{60} = a_1 - 4(60 - 1)\)

\(a^{60} = 67 - 4 \cdot 59\)

\(d^{60} = 169\)