Respuesta :

Answer:

Let two consecutive multiples of 3 be x and (x+3)

A/q,

x * (x+3) = 648

➡ x² + 3x = 648

➡ x² + 3x -648 = 0

➡ x² + 27x - 24x -648 = 0

➡ x ( x + 27 ) -24 ( x +27)

➡ ( x - 24) ( x + 27)

➡ x = 24 and x = -27

so, we take x = 24.

Required multiples of 3

➡ x = 24

➡ x +3 = 24+3 = 27.

sreedevi102

Answer:

The three consecutive multiples are 210, 216, 222

Step-by-step explanation:

3 consecutive multiples of 6 be : 6n , 6n + 6 , 6n + 12

Given sum of these 3 numbers is = 648

That is ,

            6n + 6n + 6 + 6n + 12 = 648

           18 n + 18 = 648

            18n = 648 - 18

             18 n = 630

                n = 35

Therefore the three consecutive multiples are :

6 x 35 , 6 x 35 + 6,  6 x 35 + 12

210 , 210 + 6 , 210 + 12

 210 , 216, 222

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