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Worked Examples · Example 8

Q.Find the sum of first 100 even natural numbers.

Sikkim CbseNCERTSubjective· 2mImportance★★★★★est
49% · 52/106 Questions
✓ Free question

The even naturals 2,4,6,…2,4,6,\ldots form an A.P. with a=2,d=2a=2,d=2; applying the sum-of-nn-terms formula for n=100n=100 gives 10100.

Sum of the first nn terms of an A.P.:

Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}\big[2a+(n-1)d\big]

where aa is the first term, dd is the common difference, and nn is the number of terms.

  1. The first 100 even natural numbers are 2,4,6,…,2002,4,6,\ldots,200 — an A.P. with a=2a=2, d=2d=2, n=100n=100.
  2. Substitute into the formula: S100=1002[2(2)+(100−1)(2)]S_{100} = \dfrac{100}{2}\big[2(2)+(100-1)(2)\big].
  3. Simplify inside the bracket: 2(2)=42(2)=4 and (99)(2)=198(99)(2)=198, so the bracket is 4+198=2024+198=202.
  4. Multiply: S100=50×202S_{100} = 50 \times 202.
  5. Compute: 50×202=1010050 \times 202 = 10100.
  6. Self-check using the shortcut 2+4+⋯+2n=2(1+2+⋯+n)=2⋅n(n+1)2=n(n+1)2+4+\cdots+2n = 2(1+2+\cdots+n) = 2\cdot\dfrac{n(n+1)}{2}=n(n+1): for n=100n=100, 100×101=10100100\times101=10100. ✓ Matches.
✓Final answer

The sum of the first 100 even natural numbers is 10100\mathbf{10100}.

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