Advertisements
Advertisements
प्रश्न
Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.
Advertisements
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
APPEARS IN
संबंधित प्रश्न
Find the sum of all numbers from 50 to 350 which are divisible by 6. Hence find the 15th term of that A.P.
If the sum of the first n terms of an AP is 4n − n2, what is the first term (that is, S1)? What is the sum of the first two terms? What is the second term? Similarly, find the 3rd, the 10th, and the nth terms.
Find the sum of all odd natural numbers less than 50.
Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.
How many two-digit numbers are divisible by 6?
Suppose the angles of a triangle are (a − d), a , (a + d) such that , (a + d) >a > (a − d).
In an A.P. a = 2 and d = 3, then find S12.
The famous mathematician associated with finding the sum of the first 100 natural numbers is ______.
Find the sum:
`4 - 1/n + 4 - 2/n + 4 - 3/n + ...` upto n terms
The sum of n terms of an A.P. is 3n2. The second term of this A.P. is ______.
