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प्रश्न
Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.
उत्तर
Natural numbers between 250 and 1000 which are divisible by 9 are as follows:
252, 261, 270, 279, ......, 999
Clearly, this forms an A.P. with the first term a = 252, common difference d = 9 and last term l = 999
l = a + (n – 1)d
`=>` 999 = 252 + (n – 1) × 9
`=>` 747 = (n – 1) × 9
`=>` n – 1 = 83
`=>` n = 84
Sum of first n terms = `S = n/2[a + 1]`
`=>` Sum of natural numbers between 250 and 1000 which are divisible by 9
= `84/2 [252 + 999]`
= 42 × 1251
= 52542
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Determine the sum of first 100 terms of given A.P. 12, 14, 16, 18, 20, ......
Activity :- Here, a = 12, d = `square`, n = 100, S100 = ?
Sn = `"n"/2 [square + ("n" - 1)"d"]`
S100 = `square/2 [24 + (100 - 1)"d"]`
= `50(24 + square)`
= `square`
= `square`
