Question

Find the sum of the first 15 multiples of 8.

Hint:

Basic Proportionality Theorem: If \( \frac{PS}{SQ} = \frac{PT}{TR} \), then \( ST \parallel QR \).

Answer 1

Multiples of 8 form an A.P.:
\[ 8, 16, 24, \dots \] \[ a = 8, \quad d = 8, \quad n = 15 \] \[ S_n = \frac{n}{2} [2a + (n-1)d] \] \[ S_{15} = \frac{15}{2} [2(8) + (15-1)(8)] \] \[ = \frac{15}{2} [16 + 112] = \frac{15}{2} \times 128 = 960 \] Thus, \( S_{15} = \mathbf{960} \).
Correct Answer: \( 960 \)