# the sum of three consecutive multiples of 11 is363. find these multiples

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Let us consider that the three consecutive multiples of 11 are

11n, 11(n + 1) and 11(n + 2), where n is a Natural number.

Given that,

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

⇒ 11(n + n + 1 + n + 2) = 363

⇒ 11(3n + 3) = 363

⇒ 3n + 3 = 363/11

⇒ 3n + 3 = 33

⇒ 3n = 33 - 3

⇒ 3n = 30

⇒ n = 30/3

⇒ n = 10

Thus, the required multiples of 11 are

(11 × 10), 11(10 + 1) and 11(10 + 2)

i.e., 110, 121, 132

#MarkAsBrainliest

Let us consider that the three consecutive multiples of 11 are

11n, 11(n + 1) and 11(n + 2), where n is a Natural number.

Given that,

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

⇒ 11(n + n + 1 + n + 2) = 363

⇒ 11(3n + 3) = 363

⇒ 3n + 3 = 363/11

⇒ 3n + 3 = 33

⇒ 3n = 33 - 3

⇒ 3n = 30

⇒ n = 30/3

⇒ n = 10

Thus, the required multiples of 11 are

(11 × 10), 11(10 + 1) and 11(10 + 2)

i.e., 110, 121, 132

#MarkAsBrainliest

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