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# Find the sum of all even integers between 101 and 999.

Given:

To do:

We have to find the sum of all even

Solution:

Even integers between 101 and 999 are \( 102,104,106, \ldots, 998 \).

The sequence is in A.P.

Here,

\( a=102 \) and \( d=104-102=2 \) \( l=998 \)

We know that,

$l=a+(n-1) d$

$\Rightarrow 998=102+(n-1) \times 2$

$\Rightarrow 998=102+2n-2$

$\Rightarrow 998-100=2n$

$\Rightarrow n=\frac{898}{2}=449$

$\therefore n=449$

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{449}{2}[2 \times 102+(449-1) \times 2]$

$=\frac{449}{2}[204+448 \times 2]$

$=\frac{449}{2}(1100)$

$=449 \times 550$

$=246950$

The sum of all even integers between 101 and 999 is $246950$.

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