Question

Find the sum of all natural numbers between 1 and 301 which are divisible by 11



Please answer fast

Answer 1

Answer:

4158

Step-by-step explanation:

The sum would be = 11 + 11 * 2 + 11 * 3 + . . .

Say the last number in this series = 11 * x.

So, 11 * x <= 301

=> x <= 27.36

=> x = 27

So, the required sum,

S = 11 + 11 * 2 + 11 * 3 + . . . + 11 * 27

  = 11 (1 + 2 + 3 + ... + 27)

  = 11 * (27*28)/2

  = 4158

Answer 2

Given :

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• Sum of all natural numbers between 1 and 301 which are divisible by 11.

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To find :

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• The sum of all natural numbers between 1 and 301 which are divisible by 11.

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Let's understand :

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We're asked to find the sum of all natural numbers between 1 and 301 which are divisible by 11, We'll solve this using arithmetic progression, The formula is A(n) = a + (n-1)d, Here, an = 297, As it is the last number divisible by 11, a = 11, As it the first number divisible by 11 and d is equal to 11.

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According to the question,

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So, Let's apply the formula given below and find the sum,

⇝ A(n) = a + (n-1)d

⇝ 297 = 11 + (n - 1)11

⇝ 297 - 11 = (n - 1)11

⇝ 286 = (n - 1)11

⇝ n - 1 = 286/11

⇝ n - 1 = 26

⇝ n = 26 + 1

⇝ n = 27

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Now, For sum, We need to apply the formula given below,

• Sn = n/2 ( an+a)

By applying the formula, We get,

⇝ 27/2 ( 297 + 11)

⇝ 13.5(308)

⇝ 4158

Therefore, The sum of all natural numbers between 1 and 301 which are divisible by 11 is 4158.

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