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Method and examples
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Arithmetic Progression |
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Problem 12 of 19 |
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12. Find the sum of all natural numbers between to and which are divisible by .
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Solution
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Solution provided by AtoZmath.com
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This is demo example. Please click on Find button and solution will be displayed in Solution tab (step by step)
| 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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