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Arithmetic Progression 



Problem 12 of 19 


12. Find the sum of all natural numbers between to and which are divisible by .










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Arithmetic Progression 
12. Find the sum of all natural numbers between 100 to 200 and which are divisible by 4 .
Numbers between `100` and `200` divisible by `4` are `100, 104, 108 ...`
Which are in arithmetic progression. In which `a=100` and `d=4`
Let `n` be the term such that `f(n) = 200`
We know that, `f(n) = a + (n  1)d`
`100 + (n  1)(4) = 200`
`(n  1)(4) = 100`
`n  1 = 25`
`n = 26`
We know that, `S_n = n/2 [2a + (n  1)d]`
`:. S_26 = 26/2 * [2(100) + (26  1)(4)]`
`= 13 * [200 + (100)]`
`= 13 * [300]`
`= 3900`





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