Question

if the sum of four alternate integers is 1292 then what would be the largest of these intergers

Answer 1

Answer:

326

Step-by-step explanation:

let the smallest number be X,

therefore, other numbers are, X + 2; X + 4 AND X + 6

therefore, according to question,

  X + X + 2 + X + 4 + X + 6 = 1292

= 4X + 12 = 1292

= 4X = 1292-12

= X = 1280/4 = 320

THEREFORE, LARGEST NUMBER IS X + 6 = 320 + 6 = 326