Find the sum of odd integers from 1 to 2001.

Asked by Pragya Singh | 1 year ago |  70

##### Solution :-

The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001.

It clearly forms a sequence in A.P.

Where, the first term, a = 1

Common difference, d = 2

Now,

a + (n -1)d = 2001

1 + (n-1)(2) = 2001

2n – 2 = 2000

2n = 2000 + 2 = 2002

n = 1001

We know,

Sn = $$\dfrac{n}{2}[2a+(n-1)d]$$

$$\dfrac{1001}{2}[2\times 1+(1001-1)\times 2]$$

$$\dfrac{1001}{2}[2+1000 \times 2]$$

$$\dfrac{1001}{2}\times 2002$$

= 1002001

Answered by Abhisek | 1 year ago

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