Question

the sum of three consecutive multiple of 11 is 883.find thease multiples​

Answer 1

Let the first multiple of 8 be 8x.Therefore the second consecutive multiple of 8 will be 8(x+1)Also the third consecutive multiple of 8 will be 8(x+2).It is given that the sum of these three consecutive multiples of 8 is 888=> 8x + 8(x+1) + 8(x+2) = 888=> 8x + 8x + 8 + 8x + 16 = 888=> 24x + 24 = 888Take 24 on the RHS=> 24x = 888 - 24=> x = 864/24=> x = 36.Therefore First multiple of 8 be 8x = 8 x 36 = 288Second Multiple of 8 be 8(x + 1) = 8(36 + 1) = 8 x 37 = 296Third Multiple of 8 be 8(x + 2) = 8(36 + 2) = 8 x 38 = 304If we sum up these three multiples i.e (288 + 296 + 304) we get 888.

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Answer 2

Answer:

the sum of three consecutive multiple of 11 is 883.find thease multiples