The first term of an arithmetic progression is 5 and the common difference is 3.

Sum formula: Sₙ = n/2 × [2a + (n−1)d]. Here a = 5, d = 3, so 440 = n/2 × [10 + 3(n−1)] = n/2 × (3n + 7). This gives 3n² + 7n − 880 = 0. Solving (or testing values), n = 16 satisfies: S₁₆ = 16/2 × [10 + 3(15)] = 8 × 55 = 440. ✓ (The negative root is rejected as n must be a positive integer.)