NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 5: Arithematic Progressions - Exercise 5.2
Question 8

Question. 8

Decide whether each expression can be the \(n\)th term of an AP. Justify.

(i) \(2n-3\)

(ii) \(3n^2+5\)

(iii) \(1+n+n^2\)

Answer:

(i) Yes; (ii) No; (iii) No.

Detailed Answer with Explanation:

(i) Expression: \(2n - 3\)

Step 1: Write first few terms by putting values of \(n\).

For \(n=1\): \(2(1) - 3 = -1\)

For \(n=2\): \(2(2) - 3 = 1\)

For \(n=3\): \(2(3) - 3 = 3\)

So the terms are: \(-1, 1, 3, 5, ...\)

Step 2: Find the difference between consecutive terms.

\(1 - (-1) = 2,\; 3 - 1 = 2,\; 5 - 3 = 2\)

The difference is always 2 (same each time). So it is an AP.


(ii) Expression: \(3n^2 + 5\)

Step 1: Write first few terms.

For \(n=1\): \(3(1)^2 + 5 = 8\)

For \(n=2\): \(3(2)^2 + 5 = 17\)

For \(n=3\): \(3(3)^2 + 5 = 32\)

For \(n=4\): \(3(4)^2 + 5 = 53\)

So the terms are: \(8, 17, 32, 53, ...\)

Step 2: Find the differences.

\(17 - 8 = 9,\; 32 - 17 = 15,\; 53 - 32 = 21\)

The difference keeps changing (9, 15, 21, ...). Since it is not the same, it is not an AP.


(iii) Expression: \(1 + n + n^2\)

Step 1: Write first few terms.

For \(n=1\): \(1 + 1 + 1^2 = 3\)

For \(n=2\): \(1 + 2 + 2^2 = 7\)

For \(n=3\): \(1 + 3 + 3^2 = 13\)

For \(n=4\): \(1 + 4 + 4^2 = 21\)

So the terms are: \(3, 7, 13, 21, ...\)

Step 2: Find the differences.

\(7 - 3 = 4,\; 13 - 7 = 6,\; 21 - 13 = 8\)

The difference is not the same (4, 6, 8, ...). So it is not an AP.

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 5: Arithematic Progressions – Exercise 5.2 | Detailed Answers