Question

find the sum of even positive integers between 1 and 200​

Answer 1

Answer:

Step-by-step explanation:

Let , The sum of even positive integers between 11 and 200 be S(n)S(n)

The given are the even numbers between 11 and 200 in the form of AP :

2, 4, 6 .......1982,4,6.......198

Here,

First term, a = 2a=2

Common Difference, d = a(2) - a(1) = ( 4 - 2 ) = 2d = a(2)−a(1) =(4−2)=2

$a_{n} = a + (n-1)d$

198 = 2 + (n−1)2

198 - 2 =   (n−1)2

$\frac{196}{2} = n-1$

n = 98 + 1

n = 99

We know,

$S_{n} = \frac{n}{2} (a+a_{n} )$

     =  $\frac{99}{2} (2 +198)$

      = $\frac{99}{2} (200)$

     = 99 × 100

       = 9900

Thus, the sum of even positive integers between 11 and 200 is 9900.

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Answer 2

Answer:

9900

Step-by-step explanation:

Let , The sum of even positive integers between 11 and 200 be S(n)S(n)

The given are the even numbers between 11 and 200 in the form of AP :

2, 4, 6 .......1982,4,6.......198

Here,

First term, a = 2a=2

Common Difference, d = a(2) - a(1) = ( 4 - 2 ) = 2d = a(2)−a(1) =(4−2)=2

198 = 2 + (n−1)2

198 - 2 =   (n−1)2

n = 98 + 1

n = 99

We know,

    =  

     =

    = 99 × 100

      = 9900

Thus, the sum of even positive integers between 11 and 200 is 9900.

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