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Q12:
Find the sum of the first 40 positive integers divisible by 6.

Solution :

Given: The sequence consists of the first 40 positive integers that are divisible by 6.

To Find: The sum of these first 40 positive integers.

Step 1: Identifying the Arithmetic Progression (AP)

The positive integers divisible by 6 are: $6, 12, 18, 24, \dots$

Let the sequence be represented as an Arithmetic Progression where:

  • The first term ($a$) = $6$
  • The common difference ($d$) = $12 - 6 = 6$
  • The number of terms ($n$) = $40$

Step 2: Stating the Formula for the Sum of $n$ terms

The sum of the first $n$ terms of an Arithmetic Progression is given by the formula:

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

[Where $S_n$ is the sum of $n$ terms, $a$ is the first term, $n$ is the number of terms, and $d$ is the common difference.]

Step 3: Substituting the values into the formula

Substitute $n = 40$, $a = 6$, and $d = 6$ into the formula:

$S_{40} = \frac{40}{2} [2(6) + (40 - 1)6]$

Step 4: Performing the arithmetic calculations

First, simplify the fraction outside the brackets:

$S_{40} = 20 [2(6) + (39)6]$

Next, calculate the products inside the brackets:

$S_{40} = 20 [12 + 234]$

Add the terms inside the brackets:

$S_{40} = 20 [246]$

Finally, multiply the result by 20:

$S_{40} = 4920$

Step 5: Verification (Alternative Method)

The $n^{th}$ term ($a_n$) is given by $a_n = a + (n-1)d$.

$a_{40} = 6 + (40-1)6 = 6 + 234 = 240$.

Using the alternative sum formula $S_n = \frac{n}{2}(a + l)$, where $l$ is the last term:

$S_{40} = \frac{40}{2}(6 + 240) = 20(246) = 4920$.

[Since both methods yield the same result, the calculation is verified.]

Final Answer: The sum of the first 40 positive integers divisible by 6 is 4920.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.3


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


Chapters in CBSE - Class 10 Mathematics


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