Question

what is the sum of even integers from 100 to 1000?

Answer 1

100,101..................... 1000

using an = a+(n-1)d

1000= 100+(n-1)1

n=901

using Sn =n/2(a+l)

901/2(100+1000)

=495550

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Answer 2

Answer:

the AP for even integers from 100 to1000 are:

100,102,104,106,108...........1000

therefore

a=100

d=102-100

=2

an= 1000

an=a+(n-1)d

1000 =100+(n-1)2

1000=100+2n-2

1000=98+2n

2n=1000-98

2n=902

n=902/2

n=451

Sn= n/2(a+an)

=451/2(100+1000)

=451/2(1100)

=451*550

=248050

hope this helps you^_^