How many terms of the AP −20, −16, −12, … must be taken so that the sum equals 120?

Introduction / Context:
Determining how many terms are needed to reach a target sum in an AP is a standard application of the partial-sum formula, producing a quadratic equation in n.

Given Data / Assumptions:

• a_1 = −20, d = 4.
• Find n such that S_n = 120.

Concept / Approach:
Use S_n = n/2 * [2a_1 + (n−1)d], set equal to 120, and solve for n.

Step-by-Step Solution:
S_n = n/2 * (−40 + 4(n−1)) = n/2 * (4n − 44) = n(2n − 22).Set 2n^2 − 22n = 120 ⇒ 2n^2 − 22n − 120 = 0 ⇒ n^2 − 11n − 60 = 0.Solve: (n − 15)(n + 4) = 0 ⇒ n = 15 (positive solution).

Verification / Alternative check:
Compute S_15 quickly: average term * number of terms. a_15 = −20 + 14*4 = 36; average = (−20 + 36)/2 = 8; S_15 = 15 * 8 = 120.

Why Other Options Are Wrong:
10, 20, 5 produce sums distinct from 120 (either too small or too large).

Common Pitfalls:
Dropping the factor 1/2; sign mistakes in 2a_1; forgetting S_n forms a quadratic in n.

Final Answer:
15