Question

the sum of 5 consecutive integers is 110 what is the fourth number in this sequence?

Answer 1

Answer:

The fourth number in this sequence is 23.

Step-by-step explanation:

Let the required 5 consecutive numbers are a , a + 1 , a + 2 , a + 3 , a + 4 .

According to the question :

⇒ Sum of 5 consecutive integers = 110

= >  a + ( a + 1 ) + ( a +2 ) +( a + 3 ) + ( a + 4 ) = 110

= >  a + a + 1 + a + 2 + a + 3 + a + 4 = 110

= >  a + a + a + a + a + 1 + 2 + 3 + 4 = 110

= >  5a + 10 = 110

= >  5( a + 2 ) = 110

= >  a + 2 = 22

= >  a = 22 - 2

= >  a = 20

Therefore,

Fourth number in this sequence = a + 3 = 20 + 3 = 23