Question

Find Three consecutive odd integers such that the sum of the two least integer is 11 more than the greatest integer. ??


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it's Isha​

Answer 1

Step-by-step explanation:

Let the three consecutive odd integers be 2n+1 , 2n+3, 2n+5

The two least integers = 2n+1 and 2n+3

Their sum = (2n+1)+(2n+3)

= (2n+2n)+(1+3)

= 4n+4

The greatest integer = 2n+5

According to the given problem

The sum of the two least integers is 11 more than the greatest integer.

=> 4n+4 = 2n+5+11

=> 4n+4= 2n+16

=> 4n-2n = 16-4

=> 2n = 12

=> n = 12/2

=> n = 6

If n = 6 then 2n+1 = 2(6)+1 = 12+1 = 13

If n = 6 then 2n+3 = 2(6)+3 = 12+3 = 15

If n = 6 then 2n+5 = 2(6)+5 = 12+5 = 17

Therefore,The three consecutive odd integers are 13,15,17

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Answer 2

Answer:

Let the three consecutive odd integers be 2n+1 , 2n+3, 2n+5

The two least integers = 2n+1 and 2n+3

Their sum = (2n+1)+(2n+3)

= (2n+2n)+(1+3)

= 4n+4

The greatest integer = 2n+5

According to the given problem

The sum of the two least integers is 11 more than the greatest integer.

=> 4n+4 = 2n+5+11

=> 4n+4= 2n+16

=> 4n-2n = 16-4

=> 2n = 12

=> n = 12/2

=> n = 6

If n = 6 then 2n+1 = 2(6)+1 = 12+1 = 13

If n = 6 then 2n+3 = 2(6)+3 = 12+3 = 15

If n = 6 then 2n+5 = 2(6)+5 = 12+5 = 17

Therefore,The three consecutive odd integers are 13,15,17

Step-by-step explanation:

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