NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 5: Arithematic Progressions - Exercise 5.3

Question 1

Step 1: For (A₁)

The sequence is: 2, −2, −6, −10, …

To find the common difference, subtract any term from the next term:

−2 − 2 = −4, −6 − (−2) = −4, and so on.

So, the common difference \(d = -4\).

This matches with (B₄).

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Step 2: For (A₂)

We are told: first term \(a = -18\), 10th term \(a_{10} = 0\).

General formula for the nth term of an AP is:

\(a_n = a + (n-1)d\).

So, \(a_{10} = -18 + (10-1)d = -18 + 9d\).

It is given that \(a_{10} = 0\).

So, \(0 = -18 + 9d\).

⇒ \(9d = 18\).

⇒ \(d = 2\).

This matches with (B₅).

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Step 3: For (A₃)

We are told: \(a = 0\), 10th term \(a_{10} = 6\).

Again, use the nth term formula:

\(a_{10} = a + (10-1)d = 0 + 9d = 9d\).

But we know \(a_{10} = 6\).

So, \(9d = 6\).

⇒ \(d = \dfrac{6}{9} = \dfrac{2}{3}\).

This matches with (B₁).

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Step 4: For (A₄)

We are told: second term \(a_2 = 13\), fourth term \(a_4 = 3\).

General term formula: \(a_n = a + (n-1)d\).

So, \(a_2 = a + d = 13\). … (1)

Also, \(a_4 = a + 3d = 3\). … (2)

Subtract (1) from (2):

(a + 3d) − (a + d) = 3 − 13

⇒ 2d = −10

⇒ d = −5

This matches with (B₂).

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Final Matches:

(A₁) → (B₄), (A₂) → (B₅), (A₃) → (B₁), (A₄) → (B₂).