Advertisements
Advertisements
Question
Find the sum of first 12 natural numbers each of which is a multiple of 7.
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
How many terms of the series 54, 51, 48, …. be taken so that their sum is 513 ? Explain the double answer
In an AP given a3 = 15, S10 = 125, find d and a10.
Find the sum of the first 15 terms of each of the following sequences having the nth term as
yn = 9 − 5n
Find the sum of two middle most terms of the AP `-4/3, -1 (-2)/3,..., 4 1/3.`
If in an A.P. Sn = n2p and Sm = m2p, where Sr denotes the sum of r terms of the A.P., then Sp is equal to
The given terms are 2k + 1, 3k + 3 and 5k − 1. find AP.
Find second and third terms of an A.P. whose first term is – 2 and the common difference is – 2.
Find the value of a25 – a15 for the AP: 6, 9, 12, 15, ………..
If the first term of an A.P. is p, second term is q and last term is r, then show that sum of all terms is `(q + r - 2p) xx ((p + r))/(2(q - p))`.
The sum of A.P. 4, 7, 10, 13, ........ upto 20 terms is ______.
