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S = a 1 + a 2 + a 3 + ………….. + a n−1 + a n

S = a + (a + d) + (a + 2d) + (a + 3d) + ……….. + {a + (n - 2)d} + {a + (n - 1)d}
……………….. (i)

By writing the terms of S in the reverse order, we get,

S = {a + (n - 1)d} + {a + (n - 2)d} + {a + (n - 3)d} + ……….. + (a + 3d) + (a + 2d) + (a + d)
+ a

Adding the corresponding terms of (i) and (ii), we get

2S = {2a + (n - 1)d} + {2a + (n - 1)d} + {2a + (n - 1)d} + ………. + {a + (n - 2)d}

2S = n[2a + (n -1)d

⇒ S = [2a + (n - 1)d]

Now, l = last term = nth term = a + (n - 1)d

Therefore, S = [2a + (n - 1)d] = [a {a + (n - 1)d}] = [a + l].

Sum of First n Natural Numbers

Let S be the required sum.

Therefore, S = 1 + 2 + 3 + 4 + 5 + .................... + n

Clearly, it is an Arithmetic Progression whose first term = 1, last term = n and number of
terms = n.

Therefore, S = (n + 1), [Using the formula S = (a + l)]

Example 1- Find the sum of first 25 natural numbers.

Solution- Let S be the required sum.

Therefore, S = 1 + 2 + 3 + 4 + 5 + .................... + 25

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