NCERT Solutions
Class 10 - Mathematics - Chapter 5: ARITHMETIC PROGRESSIONS - Exercise 5.1
Question 3

Question. 3

For the following APs, write the first term and the common difference:

(i) \(3, 1, -1, -3, \ldots\)

(ii) \(-5, -1, 3, 7, \ldots\)

(iii) \(\tfrac{1}{3}, \tfrac{5}{3}, \tfrac{9}{3}, \tfrac{13}{3}, \ldots\)

(iv) \(0.6, 1.7, 2.8, 3.9, \ldots\)

Answer:

(i) First term \(a = 3\), common difference \(d = -2\).

(ii) First term \(a = -5\), common difference \(d = 4\).

(iii) First term \(a = \tfrac{1}{3}\), common difference \(d = \tfrac{4}{3}\).

(iv) First term \(a = 0.6\), common difference \(d = 1.1\).

Detailed Answer with Explanation:

Recall: In an arithmetic progression (AP):

• The first term is the very first number in the list (denoted by \(a\)).
• The common difference (denoted by \(d\)) is the difference between any term and the previous term: \(d = a_2 - a_1 = a_3 - a_2 = \ldots\).

(i) \(3, 1, -1, -3, \ldots\)

First term: \(a = 3\).

Common difference: subtract consecutive terms: \(d = 1 - 3 = -2\). Check with next: \(-1 - 1 = -2\), \(-3 - (-1) = -2\). So \(d = -2\).

(ii) \(-5, -1, 3, 7, \ldots\)

First term: \(a = -5\).

Common difference: \(d = -1 - (-5) = -1 + 5 = 4\). Check with next: \(3 - (-1) = 4\), \(7 - 3 = 4\). So \(d = 4\).

(iii) \(\tfrac{1}{3}, \tfrac{5}{3}, \tfrac{9}{3}, \tfrac{13}{3}, \ldots\)

First term: \(a = \tfrac{1}{3}\).

Common difference: \(d = \tfrac{5}{3} - \tfrac{1}{3} = \tfrac{4}{3}\). Check with next: \(\tfrac{9}{3} - \tfrac{5}{3} = \tfrac{4}{3}\), \(\tfrac{13}{3} - \tfrac{9}{3} = \tfrac{4}{3}\). So \(d = \tfrac{4}{3}\).

(iv) \(0.6, 1.7, 2.8, 3.9, \ldots\)

First term: \(a = 0.6\).

Common difference: \(d = 1.7 - 0.6 = 1.1\). Check with next: \(2.8 - 1.7 = 1.1\), \(3.9 - 2.8 = 1.1\). So \(d = 1.1\).

Summary: In each case, the first term is the first number, and the common difference is obtained by subtracting any term from the next one.

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