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

Question:

Find the sum of the following APs: 1/15,1/12,1/10, . . ., to 11 terms.

Given:

An arithmetic progression (AP) with first term a = 115, second term = 112, third term = 110, and number of terms n = 11.

To Find:

The sum of the given AP up to 11 terms (S₁₁).

Formula:

The sum of an AP is given by: S_n = n2[2a + (n - 1)d], where 

'a' is the first term, 

'd' is the common difference, and 

'n' is the number of terms.

Solution:

First, we find the common difference (d): d = 112 - 115 = 5 - 460 = 160

The sum of an AP: S₁₁ = 112[2(115) + (11 - 1)(160)] 

⇒ S₁₁ = 112[215 + 1060]

⇒ S₁₁ = 112[120 + 15015x60] = 112[270900] = 112 × 310 = 3320

Result:

Therefore, the sum of the given AP up to 11 terms is equal to \( \frac{33}{20}\) 

Next question solution:

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 2(i)

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