CDe sum of three conjuctive multiple af 8 At in M.Bond Me multikelenIe

1.	CDe sum of three conjuctive multiple af 8 At in M.Bond Me multikelenIetlille
Answer» 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 = 888 Take 24 on the RHS => 24x = 888 - 24 => x = 864/24 => x = 36. Therefore First multiple of 8 be 8x = 8 x 36 = 288 Second Multiple of 8 be 8(x + 1) = 8(36 + 1) = 8 x 37 = 296 Third Multiple of 8 be 8(x + 2) = 8(36 + 2) = 8 x 38 = 304 If we sum up these three multiples i.e (288 + 296 + 304) we get 888.

CDe sum of three conjuctive multiple af 8 At in M.Bond Me multikelenIetlille

Answer»

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 = 888

Take 24 on the RHS

=> 24x = 888 - 24

=> x = 864/24

=> x = 36.

Therefore First multiple of 8 be 8x = 8 x 36 = 288

Second Multiple of 8 be 8(x + 1) = 8(36 + 1) = 8 x 37 = 296

Third Multiple of 8 be 8(x + 2) = 8(36 + 2) = 8 x 38 = 304

If we sum up these three multiples i.e (288 + 296 + 304) we get 888.