Question

Find The sum of all The even natural numbers till 1001 .
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Answer 1

To Find :

The sum of all The even number between 1 to 1001 .

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We know that:

the sequence of even number between 1-1001 is :

2,4,6,8,10,...,...,...,...,1000

Hence, This is an AP with :

• starting term = 2
• ending term = 1000
• common difference = 2

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Formula:

$\tt{ \bull \: a_n \: = a + (n - 1)d}$

where,

$\bull \tt{a_n = ending \: term}$

$\bull \tt{a = \: starting \: term}$

$\bull \tt{n = number \: of \: terms}$

$\bull \tt{d = common \: difference}$

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Applying Formula:

The above Formula is for finding the number of terms in the sequence:

given, starting term = 2

ending term = 1000

common difference = 2

$\tt{a_n = a + (n - 1)d}$

$\implies \tt{1000 = 2 + (n - 1)2}$

$\implies \tt{2n - 2 = 1000 - 2}$

$\implies \tt{2n = 998 + 2}$

$\implies \tt{n = \frac{1000}{2} }$

$\implies \tt{n = 500}$

hence, number of terms = 500

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Formula to find sum :

to find the sum of the sequence , the formula is :

$\bull \tt{s_n = \frac{n}{2} (2a + (n - 1)d}$

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Implying Formula :

$\implies \tt{s_n = \frac{n}{2} (2a + (n - 1)d)}$

$\tt{ \implies \: s_n = \frac{500}{2} (2 \times 2 + (500 - 1)2)}$

$\implies \tt{s_n \: = 250(4 + 998)}$

$\implies \tt{s_n = 250 \times 1002}$

$\red { \underline{ \boxed{ \tt{s_n = 250500}}}}$

hence, The sum of sequence = 250500

hence, the sum of even numbers till 1000

= 250500