NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 11

Question:

Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?

Given:

An arithmetic progression (AP): 3, 15, 27, 39, ...

To Find:

The term which is 132 more than the 54th term.

Formula:

The nth term of an AP is given by a_n = a + (n-1)d, where a is the first term and d is the common difference.

Solution:

First, find the common difference (d): d = 15 - 3 = 12

Find the 54th term (a₅₄): a₅₄ = 3 + (54-1)12 = 3 + 53(12) = 639

Let the required term be a_n. Then a_n = a₅₄ + 132 ⇒ a_n = 639 + 132 = 771

Now, find n: 771 = 3 + (n-1)12 ⇒ 768 = (n-1)12 ⇒ n - 1 = 76812 ⇒ n - 1 = 64 ⇒ n = 65

Therefore, the 65th term is 132 more than the 54th term.

Let's verify: a₆₅ = 3 + (65-1)12 = 771. a₅₄ = 639. 771 - 639 = 132

Result:

The 65th term of the AP is 132 more than its 54th term.

Next question solution:

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 12

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