Answer:

Three consecutive multiples are 110,121 and 132 which has the sum of 363. 33C = 363; C = 33, By substituting C = 33 in numbers 11C, 11(C – 1) and 11(C + 1) we get values 110, 121, 132.

Step-by-step explanation:

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Answer:

110, 121, 132

Step-by-step explanation:

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

So,

11x + 11(x+1) + 11(x+2) = 363

11x + 11x + 11 + 11x +22 =363

33x +33 =363

33(x + 1) =363

x + 1 = 363/33

x + 1 = 11

x = 11 - 1

x = 10

Multiples -:

• 11x = 11*10 = 110
• 11(x+1) = 11*(11) = 121
• 11(x+2) = 11*(12) = 132

VERIFICATION -: 110+121+132=> 363. HENCE PROVED.