NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 12
NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 12
Question:
Find the sum of the first 40 positive integers divisible by 6.
Given:
We need to find the sum of the first 40 positive integers divisible by 6.
To Find:
The sum of the first 40 positive integers divisible by 6.
Formula:
The sum of an arithmetic series is given by: Sn = n 2 (a1 + an), where
n is the number of terms,
a1 is the first term, and
an is the last term.
Solution:
The first 40 positive integers divisible by 6 are 6, 12, 18, ..., 240.
This is an arithmetic series with
first term a1 = 6,
common difference d = 6, and
number of terms n = 40.
The last term is a40
⇒ a40 = a1 + (n-1)d
⇒ a40 = 6 + (40-1)6
⇒ a40 = 6 + 39(6)
⇒ a40 = 240.
Using the formula for the sum of an arithmetic series: Sn = n 2 (a1 + an),
⇒ S40 = 40 2 (6 + 240)
⇒ S40 = 20(246)
⇒ S40 = 4920.
Result:
The sum of the first 40 positive integers divisible by 6 is 4920.
Next question solution:
NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 13
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