Question

sum of 44 consecutive integers is 326326.
What is the second number in this sequence?

Answer 1

Answer:

The second number of this sequence is 7395.

Step-by-step explanation:

Let the required numbers are a , a + d , a + 2d , a + 3d ..... upto 44 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.

Now,

First term = a

Common Difference = 1

Number of terms = 44

Sum of all terms = 326326

We know that the sum of n terms remains $\dfrac{n}{2}[2a+(n-1)d ]$, where  n is the number of terms, a is the first term and d is the common difference.

So,

= >  326326 = 44 / 2 [ 2( a + 1 ) + ( 44 - 1 )1 ]

= >  326326 = 22[ 2a + 2 + 43 ]

= >  326326 / 22 = 2a + 45

= >  14833 = 2a + 45

= >  14833 - 45 = 2a

= >  14788 = 2a

= >  14788 / 2 = a

= >  7394 = a

Therefore,

second number = a + d = 7394 + 1 = 7395