Question

Find the sum of odd integers from 1 to 2001.​

Answer 1

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A.P = 1, 3, 5, 7, ... , 2001.

Here,

$a = 1 \:, \: d \: = \: 2 \: , \: a_{n} \: = 2001$


Now,

$\implies \: a_{n} = \: 2001$

$\implies \: a \: + \: (n - 1)d \: = \: 2001$

$\implies \: (n - 1)2 \: = \: 2000$

$\implies \: n - 1 \: = \: 1000$

$\implies \: n \: = \: 1001$

So, we got n = 1001,


Now ,

$\implies \: S_{n} \: = \: \frac{n}{2} [ \: 2a \: + \: (n \: - 1)d ]$

$= \: \frac{1001}{2} [ \: 2(1) \: + \: (1001 \: - 1)2 ]$

$= \: \frac{10001}{2} \times 2(1 \: + \: 1000)$

$= \: 1001 \: (1001)$

$= \: 1,002,001$


Therefore, Sum of odd integers from 1 to 2001 is 1,002,001.

Answer 2

Step-by-step explanation:

Integers from 1 to 2001 are 1,2,3,4,......2001

Odd integers from 1 to 2001 are 1,3,5,......1999,2001

this sequence forms an A.P as difference between the consecutive terms is constant.

So the A.P is 1,3,5......1999,2001

Here,

First term = a = 1

Common difference = d

= 3-1

= 2

& last term = l = 2001