I can extend the pattern by adding, but for the 50th term that's too slow. Is there a direct formula and how do I use it?
Yes, use the nth term formula aₙ = a + (n − 1)d, where a is the first term, d is the common difference, and n is the term number. First find d by subtracting any term from the next one. In 5, 8, 11, 14..., the first term a = 5 and d = 8 − 5 = 3. To get the 50th term, substitute n = 50: a₅₀ = 5 + (50 − 1) × 3 = 5 + 49 × 3 = 5 + 147 = 152. The trick is (n − 1), not n, because the first term already exists before any common difference is added. Once you know a and d, you can jump straight to any term without listing them all.
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