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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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