Advertisements
Advertisements
प्रश्न
Find the sum of first 12 natural numbers each of which is a multiple of 7.
Advertisements
उत्तर
First 12 natural numbers which are multiple of 7 are as follows:
7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84
Clearly, this forms an A.P. with first term a = 7,
Common difference d = 7 and last term l = 84
Sum of first n terms = `S = n/2 [a + l]`
`=>` Sum of first 12 natural numbers which are multiple of 7
= `12/2 [7 + 84]`
= 6 × 91
= 546
APPEARS IN
संबंधित प्रश्न
In an AP given an = 4, d = 2, Sn = −14, find n and a.
The first term of an A.P. is 5, the last term is 45 and the sum of its terms is 1000. Find the number of terms and the common difference of the A.P.
How many three-digit numbers are divisible by 9?
How many terms of the AP 63, 60, 57, 54, ….. must be taken so that their sum is 693? Explain the double answer.
The nth term of an AP is given by (−4n + 15). Find the sum of first 20 terms of this AP?
In an A.P. the 10th term is 46 sum of the 5th and 7th term is 52. Find the A.P.
Find the sum of first 10 terms of the A.P.
4 + 6 + 8 + .............
How many terms of the A.P. 27, 24, 21, …, should be taken so that their sum is zero?
The sum of the first 15 multiples of 8 is ______.
If the first term of an AP is –5 and the common difference is 2, then the sum of the first 6 terms is ______.
