NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 8

Question:

An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Given:

Number of terms (n) = 50
3rd term (a₃) = 12
Last term (a₅₀) = 106

To Find:

29th term (a₂₉)

Formula:

a_n = a + (n-1)d
where a is the first term, d is the common difference, and n is the number of terms.

Solution:

a₃ = a + 2d = 12 ⇒ a = 12 - 2d

a₅₀ = a + 49d = 106

Substitute a = 12 - 2d into a₅₀ equation:
(12 - 2d) + 49d = 106 ⇒ 47d = 94 ⇒ d = 2

Substitute d = 2 into a = 12 - 2d:
a = 12 - 2(2) = 8

Now find a₂₉:
a₂₉ = a + (29-1)d = 8 + 28(2) = 8 + 56 = 64

a₂₉ = a + 28d = 8 + 28(2) = 64

Result:

The 29th term is 64.

Next question solution:

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 9

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