Question

Find three consecutive multiples of 5 whose sum is 360

Answer 1

Answer:

Let the first term be = a

n = 3

s3 = 360

d = 5

sn = n/2 (2a + (n-1)d)

s3 = 3/2 (2a + 2 (5))

360 = 3/2 (2a + 10)

240 - 10 = 2a

a = 115

Multiples = 115, 120, 125

Answer 2

Answer:

Multiple=115,120,125

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