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प्रश्न
Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.
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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
संबंधित प्रश्न
The sum of the first p, q, r terms of an A.P. are a, b, c respectively. Show that `\frac { a }{ p } (q – r) + \frac { b }{ q } (r – p) + \frac { c }{ r } (p – q) = 0`
Find the sum 2 + 4 + 6 ... + 200
In an A.P. the first term is 25, nth term is –17 and the sum of n terms is 132. Find n and the common difference.
The 7th term of the an AP is -4 and its 13th term is -16. Find the AP.
If (2p – 1), 7, 3p are in AP, find the value of p.
Write an A.P. whose first term is a and common difference is d in the following.
a = 6, d = –3
If the sum of n terms of an A.P. is 3n2 + 5n then which of its terms is 164?
Q.15
If the numbers n - 2, 4n - 1 and 5n + 2 are in AP, then the value of n is ______.
The sum of all odd integers between 2 and 100 divisible by 3 is ______.
