Question

The sum of 4 consecutive integers is 130. What is the third number in this sequence?

Answer 1

The answer is 33. Solution:

Let the nos. be x, x+1, x+2, x+3

x + x +1 + x+2 + x+3 =130

4x +6=130

4x=130-6

4x=124

x=124÷4

x=31

x+1=32

x+2=33

x+3=34

Answer 2

Answer:

The third number in this sequence is 33.

Step-by-step explanation:

Let the 4 consecutive integers are x, x+1, x+2, x+3.

It is given that the sum of 4 consecutive integers is 130.

$x+(x+1)+(x+2)+(x+3)=130$

$4x+6=130$

$4x=130-6$

$4x=124$

$x=31$

The value of x is 31.

So, 4 consecutive integers are 31, 32, 33, 34.

Therefore the third number in this sequence is 33.