Question

three consective even natural number are such that the product of the first and the third is greater than seven time the middle by 4 find the numbers

Answer 1

Answer:

6 , 8 & 10

Step-by-step explanation:

let say three consecutive even numbers are

2n-2  , 2n ,  2n+2

first number = 2(n-1)

second (middle) number = 2n

Third Number = 2(n+1)

Product of first & Third = 2(n-1)*2(n+1) = 4(n²-1)

product of the first and the third is greater than seven time the middle by 4

4(n²-1) = 7 * 2n  + 4

=> 2(n²-1) = 7n + 2

=> 2n² - 2 = 7n + 2

=> 2n² - 7n - 4 = 0

=> 2n² - 8n + n - 4 = 0

=> 2n(n-4) + 1(n-4) = 0

=> (2n +1)(n-4) = 0

n = -1/2 (not possible as n should be integer)

n = 4

2(n-1) , 2n , 2(n+1)

6 , 8 , 10