Arithmetic progression — first 15 multiples of 8 Compute the sum of 8, 16, 24, … up to the 15th multiple of 8.

Introduction / Context:
Sums of multiples are AP sums. Recognizing the structure allows using a compact formula to avoid enumerating every term. Here the common difference is 8 and we want the first 15 terms.

Given Data / Assumptions:

• First term a = 8.
• Common difference d = 8.
• Number of terms n = 15.

Concept / Approach:
The n-th multiple of 8 is 8n, and the sum of the first n multiples of 8 equals 8*(1 + 2 + … + n) = 8 * (n*(n+1)/2). Using this relation is faster and less error-prone than listing terms.

Step-by-Step Solution:
Compute T = 1 + 2 + … + 15 = 15*16/2 = 120.Multiply by 8: S = 8 * 120 = 960.Hence, the sum is 960.

Verification / Alternative check:
AP formula: S = n/2 * (2a + (n-1)d) = 15/2 * (16 + 14*8) = 15/2 * (16 + 112) = 15/2 * 128 = 15 * 64 = 960, confirming the value.

Why Other Options Are Wrong:
660, 1060, 1200 reflect arithmetic errors such as wrong n, misapplied formula, or skipping the /2 step.

Common Pitfalls:
Confusing the number of terms with the last term value; forgetting the factor 1/2 in the AP formula; or miscomputing 15*16.

Final Answer:
960