Question 36 of 44intermediate🔧 ApplyShort Answer3 marks

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

Correct Answer

The first 40 positive integers divisible by 6 are 6, 12, 18, 24, ... These numbers form an Arithmetic Progression (AP) with the first term a=6a = 6 and common difference d=6d = 6. We need to find the sum of the first 40 terms (n=40n = 40).

Using the formula for the sum of the first nn terms of an AP, Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d], we substitute the values:

S40=402[2(6)+(40−1)6]S_{40} = \frac{40}{2}[2(6) + (40 - 1)6] S40=20[12+39(6)]S_{40} = 20[12 + 39(6)] S40=20[12+234]S_{40} = 20[12 + 234] S40=20[246]S_{40} = 20[246] S40=4920S_{40} = 4920

Therefore, the sum of the first 40 positive integers divisible by 6 is 4920.

Exercise: EXERCISE 5.3 | Q: 12 | (Page Page 21)
For More Understanding

Explanation

The problem is solved by identifying the sequence of numbers divisible by 6 as an Arithmetic Progression (AP). The first term is 6, and the common difference is also 6. Using the standard formula for the sum of an AP provided in the context (Example 15), Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d], the values are substituted to find the final answer.

Solution Steps

  1. Step 1: Identify the sequence. The integers are 6, 12, 18, ... This is an AP with a=6a = 6 and d=6d = 6.

  2. Step 2: Identify the formula. Use the sum formula Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d].

  3. Step 3: Substitute values n=40n = 40, a=6a = 6, d=6d = 6 into the formula.

  4. Step 4: Calculate: S40=20[12+234]=20[246]=4920S_{40} = 20[12 + 234] = 20[246] = 4920.

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