Q. 23.8( 6 Votes )

Let’s write down three consecutive integers such that 5 less than the sum of the number is equal to 11 more than twice the second number. Let’s find the three consecutive integers.

Answer :

Let the three consecutive numbers be x, (x+1), (x+2).

Now according to the given criteria,

5 less than the sum of the number = x+(x+1)+(x+2)-5…..(i)

11 more than twice the second number = 2(x+1)+11…….(ii)

So from question equation (i) and equation (ii) are equal,


x+(x+1)+(x+2)-5 = 2(x+1)+11







Hence the three consecutive numbers are

15, 16 and 17

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