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,

Hence

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

⇒ 3x+3-5=2x+2+11

⇒ 3x-2=2x+13

⇒ 3x-2x=13+2

⇒ x=15

⇒ x+1=15+1=16

⇒ x+2=15+2=17

Hence the three consecutive numbers are

15, 16 and 17

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