【】 The sum of 5 consecutive integers is 110 .What is the fourth number in this sequence ?

Solution:
The fourth number of this sequence is 23.

Explanation:

We need to find out the  fourth  number  from  consecutive integer numbers' sequence.

 

                      Step-1:

We have to have knowledge about integer number.An integer number does not have any fractional part.It can be positive or negative number.

Now you have to understand about consecutive integer number.The difference between two consecutive integer numbers  is always 1.

 

                Step-2:

Say,the first integer number is x.

So, the second integer number will be x+ 1. Thus, the third consecutive integer number will be x+2 .Besides,the fourth integer number will be x+3 and the fifth number will be x+4 .

Have a look here.

The five consecutive integer numbers are x, x+1,x+2 ,x+3 ,x+4 respectively.

                  Step-3:

According to the conditions,the sum of 5 consecutive integer numbers will be 110.

 

So, we are getting the following equation

 

x +( x+1)+(x+2) +(x+3) +(x+4)  = 110

Or, 5x + 10 = 110

Or, 5x + 10 -10  = 110 - 10[ Deduct 10 from both sides ]

Or, 5x + 0 = 100

Or, 5x = 100

Or, 5x/5 =  100/5

[Divide both sides by 5]

Or, x = 20

Now put the value of x here:

   x,     x+1,  x+2 , x+3  , x+4

= 20,  20+1 , 20+2 , 20+3,20 +4

20 ,   21,      22,           23,    24

So, the fourth number of this sequence is 23.

Answer: 23

Next