English

Find the Sum of All Natural Numbers Between 1 and 100, Which Are Divisible by 3.

Advertisements
Advertisements

Question

Find the sum of all natural numbers between 1 and 100, which are divisible by 3.

Advertisements

Solution

In this problem, we need to find the sum of all the multiples of 3 lying between 1 and 100.

So, we know that the first multiple of 3 after 1 is 3 and the last multiple of 3 before 100 is 99.

Also, all these terms will form an A.P. with the common difference of 3.

So here,

First term (a) = 3

Last term (l) = 99

Common difference (d) = 3

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

`a_n = a + (n - 1)d`

So, for the last term,

99 = 3 + (n - 1)3

99 = 3 + 3n - 3

99 = 3n

Further simplifying 

`n = 99/3`

n = 33

Now, using the formula for the sum of n terms,

`S_n = 33/2 [2(3) + (33 - 1)3]`

`= 33/2 [6 + (32)3]`

`= 33/2 (6 + 96)`

`= (33(102))/2`

On further simplification, we get,

`S_n = 33(51)`

= 1683

Therefore, the sum of all the multiples of 3 lying between 1 and 100 is `S_n = 1683`

shaalaa.com
  Is there an error in this question or solution?

RELATED QUESTIONS

How many terms of the A.P. 18, 16, 14, .... be taken so that their sum is zero?


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 the following APs.

0.6, 1.7, 2.8, …….., to 100 terms. 


Find the sum of first 40 positive integers divisible by 6.


Find the sum of first 12 natural numbers each of which is a multiple of 7.


Find the sum of all multiples of 7 lying between 300 and 700.


Which term of the AP 21, 18, 15, …… is -81?


If the 10th  term of an AP is 52 and 17th  term is 20 more than its 13th  term, find the AP


The 7th term of the an AP is -4 and its 13th term is -16. Find the AP.


The first term of an AP is p and its common difference is q. Find its 10th term. 


For what value of n, the nth terms of the arithmetic progressions 63, 65, 67, ... and 3, 10, 17, ... equal?


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 Sn − kSn−1 + Sn−2, then k =


If 18, ab, −3 are in A.P., the a + b =


The common difference of the A.P.

\[\frac{1}{3}, \frac{1 - 3b}{3}, \frac{1 - 6b}{3}, . . .\] is 
 

The sum of the first 2n terms of the AP: 2, 5, 8, …. is equal to sum of the first n terms of the AP: 57, 59, 61, … then n is equal to ______.


If the numbers n - 2, 4n - 1 and 5n + 2 are in AP, then the value of n is ______.


Find the sum:

`(a - b)/(a + b) + (3a - 2b)/(a + b) + (5a - 3b)/(a + b) +` ... to 11 terms


How many terms of the AP: –15, –13, –11,... are needed to make the sum –55? Explain the reason for double answer.


The 5th term and the 9th term of an Arithmetic Progression are 4 and – 12 respectively.

Find:

  1. the first term
  2. common difference
  3. sum of 16 terms of the AP.

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×