Question

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

Answer 1

Question:-

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

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➳ Positive integer which are divisible by 6 are 6 , 12 , 18 , 24 and so on.

Given:-

• a = 6
• d = 6

To find:-

• S40 = ?

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$\large➻s_{40} = \frac{40}{2} (2 \times 6 + (40 - 1)6)$

$\large➻s_{40} = 20(12 + 39 \times 6)$

$\large➻s_{40} = 20(12 + 234)$

$\large➻s_{40} = 20 \times 246$

$\large➻s_{40} = 4920$

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✔ So the sum of first 40 terms divisible by 6 is 4920.