Advertisements
Advertisements
Question
Find the sum of first 12 natural numbers each of which is a multiple of 7.
Advertisements
Solution
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
RELATED QUESTIONS
If Sn denotes the sum of first n terms of an A.P., prove that S30 = 3[S20 − S10]
Find the sum of the following arithmetic progressions
`(x - y)^2,(x^2 + y^2), (x + y)^2,.... to n term`
Find the sum of all odd natural numbers less than 50.
Choose the correct alternative answer for the following question .
15, 10, 5,... In this A.P sum of first 10 terms is...
The first and the last terms of an A.P. are 8 and 350 respectively. If its common difference is 9, how many terms are there and what is their sum?
Let Sn denote the sum of n terms of an A.P. whose first term is a. If the common difference d is given by d = Sn − kSn−1 + Sn−2, then k =
If the sums of n terms of two arithmetic progressions are in the ratio \[\frac{3n + 5}{5n - 7}\] , then their nth terms are in the ratio
Q.5
Q.19
Find the sum of three-digit natural numbers, which are divisible by 4.
