Question

he sum of 55 consecutive integers is 120120.
What is the third number in this sequence?

Answer 1

Answer:

The third number of this sequence is 2158 .

Step-by-step explanation:

Let the required numbers are a , a + d , a + 2d , a + 3d ..... upto 55 terms.

It is given that the numbers are consecutive, so value of d should be 1.

Now,

Numbers are a , a + 1 , a + 2 , a + 4 , a + 5 ..... upto 55 terms.

Then,

First term = a

Common Difference = 1

Number of terms = 55

Sum of all terms = 120120

We know that the sum of n terms remains , where  n is the number of terms, a is the first term and d is the common difference.

So,

= >  120120 = ( 55 / 2 ) [ 2( a + 1 ) + ( 55 - 1 )1 ]

= >  120120 x 2 / 55 = 2a + 2 + 54

= >  4368 = 2a + 54

= >  4368 - 54 = 2a

= >  4314 = 2a

= >  4314 / 2 = a

= >  2157 = a

Therefore,

Third number of this sequence = a + d = 2157 + 1 = 2158.