Question

The sum of 44 consecutive integers is 198.

Answer 1

Answer:

The required sequence is - 17 , - 16 , - 15 , - 14 ....

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 44 terms.

Then,

First term = a

Common Difference = 1

Number of terms = 44

Sum of all terms = 198

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,

= >  198 = ( 44 / 2 ) [ 2( a + 1 ) + ( 44 - 1 )1 ]

= >  198 = 22[ 2a + 2 + 43 ]

= >  198 / 22 = 2a + 45

= >  9 = 2a + 45

= >  9 - 45 = 2a

= >  - 36 = 2a

= >  - 36 / 2 = a

= >  - 18 = a

Therefore,

Required sequence is a , a + 1 , a + 2 , a + 3 , a + 4 .....upto 44 terms is - 17 , - 16 , - 15 , -14 .....