Working the sum backwards

How many terms of 5, 8, 11, … add up to 75? The total is known. The count of terms, n, is missing.

Six terms add up to 75. Here is the working.

Here a = 5 is the first term. The step is d = 3. Use the other face of the sum, Sₙ = n/2 (2a + (n − 1)d). Make it equal 75.

n/2 (2 × 5 + (n − 1) × 3) = n/2 (3n + 7), so n/2 (3n + 7) = 75.

Multiply both sides by 2: n(3n + 7) = 150. That simplifies to 3n² + 7n − 150 = 0.

With a and d known, unknown n gives a quadratic. That needs d not 0. Keep only a whole count of terms, at least 1.

Factorising gives (n − 6)(3n + 25) = 0. The roots are the values of n that solve it. They are n = 6 and n = −25/3. A count of terms is a whole number, at least 1. So −25/3 is refused. Check forwards: 5 + 8 + 11 + 14 + 17 + 20 = 75. Six terms match.

A second kind of question keeps n known, and asks for d. A list starts at 12. Its first 8 terms total 40:

8/2 (2 × 12 + 7d) = 40 gives 4(24 + 7d) = 40, so 96 + 28d = 40. Then 28d = −56, so d = −2. That is one linear equation, no quadratic, because n was already given. Check: 12, 10, 8, 6, 4, 2, 0, −2 total 40.

Tap the right equation. Then tap the root a count can keep.