NCERT Solutions
Class 10 - Mathematics - Chapter 5: ARITHMETIC PROGRESSIONS - Exercise 5.2

Question 2

Formula used: For an AP with first term \(a\) and common difference \(d\), the \(n\)th term is

\[a_n = a + (n - 1)d\]

(i) 30th term of the AP: 10, 7, 4, \ldots

First term: \(a = 10\).

Common difference: \(d = 7 - 10 = -3\). Check: \(4 - 7 = -3\), so it is an AP with \(d = -3\).

We need the 30th term, i.e. \(a_{30}\):

\[a_{30} = a + (30 - 1)d = 10 + 29(-3)\]

\[a_{30} = 10 - 87 = -77\]

So the 30th term is \(-77\), which matches Option (C).

(ii) 11th term of the AP: -3, -\(\tfrac{1}{2}\), 2, \ldots

First term: \(a = -3\).

Find the common difference:

\[d = a_2 - a_1 = -\tfrac{1}{2} - (-3) = -\tfrac{1}{2} + 3 = -\tfrac{1}{2} + \tfrac{6}{2} = \tfrac{5}{2}\]

Check with next term: \(2 - (-\tfrac{1}{2}) = 2 + \tfrac{1}{2} = \tfrac{5}{2}\), so \(d = \tfrac{5}{2}\).

We need the 11th term, \(a_{11}\):

\[a_{11} = a + (11 - 1)d = -3 + 10 \cdot \tfrac{5}{2}\]

\[a_{11} = -3 + \tfrac{50}{2} = -3 + 25 = 22\]

So the 11th term is \(22\), which matches Option (B).