Advertisements
Advertisements
Question
Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.
Solution
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
RELATED QUESTIONS
Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms is 40.
The sum of the first n terms of an AP in `((5n^2)/2 + (3n)/2)`.Find its nth term and the 20th term of this AP.
Find the sum of all multiples of 9 lying between 300 and 700.
In an A.P. the first term is 8, nth term is 33 and the sum to first n terms is 123. Find n and d, the common differences.
If \[\frac{1}{x + 2}, \frac{1}{x + 3}, \frac{1}{x + 5}\] are in A.P. Then, x =
The common difference of an A.P., the sum of whose n terms is Sn, is
The common difference of the A.P.
The 5th term and the 9th term of an Arithmetic Progression are 4 and – 12 respectively.
Find:
- the first term
- common difference
- sum of 16 terms of the AP.
In a ‘Mahila Bachat Gat’, Kavita invested from the first day of month ₹ 20 on first day, ₹ 40 on second day and ₹ 60 on third day. If she saves like this, then what would be her total savings in the month of February 2020?
The nth term of an A.P. is 6n + 4. The sum of its first 2 terms is ______.
