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
In an AP Given a12 = 37, d = 3, find a and S12.
Find the sum of all 3-digit natural numbers, which are multiples of 11.
Which term of the AP 21, 18, 15, ... is –81?
The 4th term of an AP is zero. Prove that its 25th term is triple its 11th term.
If the seventh term of an A.P. is \[\frac{1}{9}\] and its ninth term is \[\frac{1}{7}\] , find its (63)rd term.
Find where 0 (zero) is a term of the A.P. 40, 37, 34, 31, ..... .
If the sum of first p term of an A.P. is ap2 + bp, find its common difference.
The common difference of the A.P. \[\frac{1}{3}, \frac{1 - 3b}{3}, \frac{1 - 6b}{3}, ...\] is ______.
Determine the sum of first 100 terms of given A.P. 12, 14, 16, 18, 20,......
Activity :- Here, a = 12, d = `square`, n = 100, S100 = ?
Sn = `"n"/2 [square + ("n" - 1)"d"]`
S100 = `square/2 [24 + (100 - 1)"d"]`
= `50(24 + square)`
= `square`
= `square`
