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11. Arithmetic Progression
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Q18 of 178 Page 551

Find the sum of first 100 even natural numbers which are divisible by 5.

First 100 even natural numbers which are divisible by 5 are 10, 20, 30, …, 1000

Here, first term = a = 10


Common difference = d = 10


Number of terms = 100


Now, Sum of n terms of this arithmetic series is given by:


Sn = [2a + (n - 1)d]


Therefore sum of 100 terms of this arithmetic series is given by:


∴ S100 = [2(10) + (100 - 1)(10)]


= 50 × [20 + 990]


= 50 × 1010


= 50500


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Questions · 178
11. Arithmetic Progression
1 1 1 1 1 2 2 2 2 2 3 4 5 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 1 1 1 1 1 2 2 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
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