Question

the sum of three consecutive multiple of 11 is 363 find these multiples

Answer 1

Let three consecutive multiple of 11 be 11n, 11(n+1), 11(n+2)
Since sum is 363 so
11n+11(n+1)+11(n+2)=363
11n+11n+11+11n+22=363
33n=330
n=10
So the multiple are 11*10=110, 11(10+1)=121, 11(10+2)=132

Answer 2

Hey!

let the three consecutive
multiples be 11x, 11(x+1), 11(x+2)

11x + 11(x+1)+11(x+2 )= 363
11x+11x+11x+11+22 = 363
33x + 33=363
33x =363-33
=330
x =330/33= 10

11n= 110
11(n+1)=110+11
=121
11(n+2)=110+22
=132

So the required multiples of 11 are-
110, 121, 132.

[110+121+132=363]

hope it helped u ^-^