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Algebra II: Sum of n terms of an A.P.
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Sum of n terms of an A.P.
The sum Sn of n terms of an A.P. with first term 'a' and common difference 'd' is 
     Sn = n/2[2a + (n - 1)d]
or, Sn = n/2[a + l], where l = last term = a + (n - 1)d
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Example: Find the sum of all three digit natural numbers, which are divisible by 5.
Solution: The smallest and largest three digit numbers, which are divisible by 5 are 100 and 995 respectively.
Sequence of the numbers divisible by 5 are 100, 105, 110,.....995. Clearly, it is an A.P. with a = 100 and d = 5.

Let the number of terms in the sequence be n. Then,

a + (n - 1)d = 995
100 + (n - 1)5 = 995
n = 180

Required Sum= n/2[2a + (n - 1)d]
= 180/2[2 x 100 + (180 - 1)5]
= 90[200 + 179 x 5]
= 98550
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