Question

Find the sum of first 15 multiples of 8.

Answer 1

The multiples of 8 are

8, 16, 24, 32, ...

These are in an A.P., having the first term as 8 and common difference as 8.

Therefore, a = 8

d = 8

S₁₅ = ?

`S_n = n/2[2a + (n - 1)d]`

= `15/2[2(8) + (15 - 1)8]`

= `15/2[16 + 14(8)]`

= `15/2(16 + 112)`

= `(15(128))/2`

= 15 × 64

= 960

Therefore, the sum of the first 15 multiples of 8 is 960.

Answer 2

Multiples of 8 are: 8, 16, 24, 32, ..........., which form an A.P. with first term, a = 8 and common difference, d = 8

∵ Sum of n^th term of A.P.

S_n = `n/2[2a + (n - 1)d]`

∴ S₁₅ = `15/2 [2 xx 8 + (15 - 1) xx 8]`

= `15/2 [16 + 112]`

= `15/2 xx 128`

= 15 × 64

= 960