NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 1(ii)

Question:

Find the sum of the following APs: –37, –33, –29, . . ., to 12 terms.

Given:

An arithmetic progression (AP): –37, –33, –29, . . . Number of terms (n) = 12

To Find:

The sum of the arithmetic progression (S_n)

Formula:

The sum of an arithmetic progression is given by: S_n = n 2 [2a + (n – 1)d] , where 'n' is the number of terms,
'a' is the first term, and
'd' is the common difference.

Solution:

Here, a = –37, d = –33 – (–37) = 4, and n = 12.

Substituting these values into the formula:
⇒ S₁₂ = 12 2 [2(–37) + (12 – 1)4]

⇒ S₁₂ = 6[–74 + 44]

⇒ S₁₂ = 6(–30)

⇒ S₁₂ = –180

Result:

The sum of the AP is -180.

Next question solution:

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 1(iii)

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