Question

A,B,C,D are 4consecutive number.if the sum of first and last numbers is 257,then product of remains two numbers is​

Answer 1

Answer:

16512

Step-by-step explanation:

A,B,C,D are 4 consecutive number so,

let A be x then B = (x+1) , C = (x+2), D = (x + 3)

so ATP, sum of first and last = 257

           A + D = 257

     x + (x+3) = 257

   2x + 3 = 257

 2x = 257-3 = 254

x = 254/2 = 127

then, B = (x + 1 )  = (127 + 1) = 128

   C = (x +2 ) = (127 + 2) = 129

products of B and C = BC = 128*129 = 16512

hope it will helps u

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Answer 2

let the 4 consecutive no be x-1,x,x+1,x+2

sum of first and last =257

x-1+x+2=257

2x+1=257

2x=256

x=128

first no = 128-1=127

second no = 128

third no =128+1=129

fourth no =128+2=130

The product of remain two= 128×129

= 16512

hope u got it