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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