Advertisements
Advertisements
Question
Find the sum of all natural numbers between 200 and 400 which are divisible by 7.
Solution
Natural numbers between 200 and 400 which are divisible by 7 are 203, 210 ,.... 399.
This is an AP with a = 203, d = 7 and l = 399.
Suppose there are n terms in the AP. Then,
an = 399
⇒ 203 + (n-1) × 7 = 399 [an = a+ (n-1) d]
⇒ 7n + 196 =399
⇒ 7n = 399-196 =203
⇒ n = 29
`∴ "Required sum " = 29/2 (203+399) [ s_n = n/2 (a+l)]`
`= 29/2 xx 602`
= 8729
Hence, the required sum is 8729.
APPEARS IN
RELATED QUESTIONS
If the sum of the first n terms of an A.P. is `1/2`(3n2 +7n), then find its nth term. Hence write its 20th term.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Find the sum of the following arithmetic progressions:
3, 9/2, 6, 15/2, ... to 25 terms
Find the sum of all odd numbers between 100 and 200.
Find the sum of the first 15 terms of each of the following sequences having the nth term as
`a_n = 3 + 4n`
The 4th term of an AP is 11. The sum of the 5th and 7th terms of this AP is 34. Find its common difference
If 4 times the 4th term of an AP is equal to 18 times its 18th term then find its 22nd term.
If the numbers (2n – 1), (3n+2) and (6n -1) are in AP, find the value of n and the numbers
Write the common difference of an A.P. whose nth term is an = 3n + 7.
Find the sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.
