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in Arithmetic Progression by (44.5k points)
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Find the sum of all odd integers from 1 to 2001.

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The odd numbers lying between 1 and 2001 are

1, 3, 5,…, 2001

a2 – a1 = 3 – 1 = 2

a3 – a2 = 5 – 3 = 2

∵ a3 – a2 = a2 – a1 = 2

Therefore, the series is in AP

Here, a = 1, d = 2 and an = 2001

We know that,

an = a + (n – 1)d

⇒ 2001 = 1 + (n – 1)2

⇒ 2001 – 1 = (n – 1)2

⇒ 2000 = (n – 1)2

⇒ 1000 = (n – 1)

⇒ n = 1001

Now, we have to find the sum of this AP

⇒ S1001 = 1001[1 + 1000]

⇒ S1001 = 1001 [1001]

⇒ S1001 = 1002001

Hence, the sum of all odd numbers lying between 1 and 2001 is 1002001.

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