Question

Find the sum of the first 20 positive integers divisible by 6

Answer 1

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Answer 2

The sum of the first 20 positive integers divisible by 6 is 1260

Given :

The first 20 positive integers divisible by 6

To find :

The sum of the first 20 positive integers divisible by 6

Formula :

Sum of first n terms of an arithmetic progression

$\displaystyle \sf S_n= \frac{n}{2} \bigg[2a + (n - 1)d \bigg]$

Where First term = a

Common Difference = d

Solution :

Step 1 of 3 :

Write down the given AP

The first 20 positive integers divisible by 6 are 6 , 12 , 18 , . . . , 120

Step 2 of 3 :

Write down first term and common difference

First term = a = 6

Common Difference = d = 12 - 6 = 6

Step 3 of 3 :

Calculate sum of the first 20 positive integers divisible by 6

Number of terms = n = 20

∴ The sum of the first 20 positive integers divisible by 6

$\displaystyle \sf = \frac{n}{2} \bigg[2a + (n - 1)d \bigg]$

$\displaystyle \sf = \frac{20}{2} \bigg[(2 \times 6) + (20 - 1) \times 6 \bigg]$

$\displaystyle \sf = 10 \times \bigg[12 + 114 \bigg]$

$\displaystyle \sf = 10 \times 126$

$\displaystyle \sf = 1260$

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