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.