NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 5: Arithematic Progressions - Exercise 5.3
Question 2

Question. 2

Verify each is an AP and write its next three terms.

(i) \(0,\; \dfrac{1}{4},\; \dfrac{1}{2},\; \dfrac{3}{4},\ldots\)

(ii) \(5,\; \dfrac{14}{3},\; \dfrac{13}{3},\; 4,\ldots\)

(iii) \(\sqrt{3},\; 2\sqrt{3},\; 3\sqrt{3},\ldots\)

(iv) \(a+b,\; (a+1)+b,\; (a+1)+(b+1),\ldots\)

(v) \(a,\; 2a+1,\; 3a+2,\; 4a+3,\ldots\)

Answer:

(i) Yes, \(d=\dfrac{1}{4}\). Next: \(1,\; \dfrac{5}{4},\; \dfrac{3}{2}\).

(ii) Yes, \(d=-\dfrac{1}{3}\). Next: \(\dfrac{11}{3},\; \dfrac{10}{3},\; 3\).

(iii) Yes, \(d=\sqrt{3}\). Next: \(4\sqrt{3},\;5\sqrt{3},\;6\sqrt{3}\).

(iv) Yes, \(d=1\). Next: \(a+b+3,\; a+b+4,\; a+b+5\).

(v) Yes, \(d=a+1\). Next: \(5a+4,\; 6a+5,\; 7a+6\).

Detailed Answer with Explanation:

Step 1: Recall the rule of an Arithmetic Progression (AP)

A sequence is called an AP if the difference between any two consecutive terms is always the same. This difference is called the common difference (d).

(i)

Terms: \(0, \; \tfrac{1}{4}, \; \tfrac{1}{2}, \; \tfrac{3}{4}\)

Find differences:

  • \(\tfrac{1}{4} - 0 = \tfrac{1}{4}\)
  • \(\tfrac{1}{2} - \tfrac{1}{4} = \tfrac{1}{4}\)
  • \(\tfrac{3}{4} - \tfrac{1}{2} = \tfrac{1}{4}\)

Since all differences are equal, it is an AP with \(d = \tfrac{1}{4}\).

Next three terms: add \(\tfrac{1}{4}\) each time → \(1, \tfrac{5}{4}, \tfrac{3}{2}\).

(ii)

Terms: \(5, \; \tfrac{14}{3}, \; \tfrac{13}{3}, \; 4\)

Find differences:

  • \(\tfrac{14}{3} - 5 = \tfrac{14}{3} - \tfrac{15}{3} = -\tfrac{1}{3}\)
  • \(\tfrac{13}{3} - \tfrac{14}{3} = -\tfrac{1}{3}\)
  • \(4 - \tfrac{13}{3} = \tfrac{12}{3} - \tfrac{13}{3} = -\tfrac{1}{3}\)

All differences are \(-\tfrac{1}{3}\), so it is an AP with \(d = -\tfrac{1}{3}\).

Next three terms: add \(-\tfrac{1}{3}\) each time → \(\tfrac{11}{3}, \tfrac{10}{3}, 3\).

(iii)

Terms: \(\sqrt{3}, \; 2\sqrt{3}, \; 3\sqrt{3}\)

Find differences:

  • \(2\sqrt{3} - \sqrt{3} = \sqrt{3}\)
  • \(3\sqrt{3} - 2\sqrt{3} = \sqrt{3}\)

All differences are equal to \(\sqrt{3}\), so it is an AP with \(d = \sqrt{3}\).

Next three terms: add \(\sqrt{3}\) each time → \(4\sqrt{3}, 5\sqrt{3}, 6\sqrt{3}\).

(iv)

Terms: \(a+b, \; (a+1)+b, \; (a+1)+(b+1)\)

Write them clearly:

  • First term: \(a+b\)
  • Second term: \(a+1+b = a+b+1\)
  • Third term: \(a+1+b+1 = a+b+2\)

Differences:

  • Second – First = \((a+b+1) - (a+b) = 1\)
  • Third – Second = \((a+b+2) - (a+b+1) = 1\)

So the common difference \(d = 1\).

Next three terms: \(a+b+3, a+b+4, a+b+5\).

(v)

Terms: \(a, \; 2a+1, \; 3a+2, \; 4a+3\)

Differences:

  • Second – First = \((2a+1) - a = a+1\)
  • Third – Second = \((3a+2) - (2a+1) = a+1\)
  • Fourth – Third = \((4a+3) - (3a+2) = a+1\)

All are equal, so common difference \(d = a+1\).

Next three terms: add \(a+1\) each time → \(5a+4, 6a+5, 7a+6\).

Conclusion: In each case, the differences are constant, so they are APs and the next terms are found by repeatedly adding the common difference.

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NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 5: Arithematic Progressions – Exercise 5.3 | Detailed Answers