1 markMCQSection A

The common difference of the AP : $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}$ , ..... is : (a) √2 (b) 1 (c) 2√2 (d) -√2

Arithmetic Progressions

Common difference of an AP

Options

(A)

\sqrt{2}

(B)1

(C)

2\sqrt{2}

(D)

-\sqrt{2}

Official Answer

(a) $\sqrt{2}$ — $d = a_2 - a_1 = 2\sqrt{2} - \sqrt{2} = \sqrt{2}$ , confirmed constant across all consecutive terms.

common differencearithmetic progressiond = a2 - a12√2 - √2√2multiples of √2constant difference

Marking Scheme

11 mark: correct option (a) $\sqrt{2}$ .
2Expected reasoning: $d = a_2 - a_1 = 2\sqrt{2} - \sqrt{2} = \sqrt{2}$ , verified as constant across the sequence.

Hint

Subtract the first term from the second: $2\sqrt{2} - \sqrt{2}$ .

Quick Oral Answer

The common difference is the second term minus the first: $2\sqrt{2} - \sqrt{2} = \sqrt{2}$ , and it stays constant throughout the sequence.

Analysis & Explanation

This MCQ checks the method for finding the common difference of an AP with surd terms.

Concept

Common difference $d = a_{n+1} - a_n$ (any term minus its preceding term).
For $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \ldots, d = 2\sqrt{2} - \sqrt{2} = \sqrt{2}$ .

Key points

Verified consistent: $3\sqrt{2} - 2\sqrt{2} = \sqrt{2}$ and $4\sqrt{2} - 3\sqrt{2} = \sqrt{2}$ .
Terms are $\sqrt{2} \times 1, \sqrt{2} \times 2, \sqrt{2} \times 3$ , ... so √2 is the natural step size.

Common mistakes

Reading off the coefficients 1,2,3,4 as the difference (giving 1) and ignoring the common factor √2.
Picking $2\sqrt{2}$ (the second term, not the difference) or the wrong sign $-\sqrt{2}$ for an increasing sequence.

Common Mistakes

1Reading off the second term $2\sqrt{2}$ as the common difference instead of subtracting consecutive terms.
2Ignoring the $\sqrt{2}$ factor and using the coefficients 1, 2, 3, 4 to get a difference of 1.
3Assigning a negative sign ( $-\sqrt{2}$ ) even though the terms are increasing, so the difference is positive.

Interesting Facts

An AP can have an irrational common difference like $\sqrt{2}$ ; the definition only requires the difference between consecutive terms to be CONSTANT, not that it be a whole number.

This AP is just $\sqrt{2}$ times the natural-number sequence 1, 2, 3, 4, ..., so its nth term is $a_n = n\sqrt{2}$ — scaling a simple AP by a constant keeps it an AP.

The concept of arithmetic progressions dates back over a thousand years; the young Carl Friedrich Gauss famously summed the AP $1+2+\ldots+100 = 5050$ in seconds as a schoolboy.

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Frequently Asked Questions

How do you find the common difference of an AP?

Subtract any term from the term that follows it: $d = a_{n+1} - a_n$ . For $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, \ldots, d = 2\sqrt{2} - \sqrt{2} = \sqrt{2}$ , and you can verify it is the same for every consecutive pair.

Can the common difference of an AP be irrational?

Yes. The only requirement for an AP is that the difference between consecutive terms is constant. That constant can be irrational, like $\sqrt{2}$ , or negative, or a fraction — it just has to stay the same throughout.

Why isn't the common difference $2\sqrt{2}$ here?

$2\sqrt{2}$ is the second term of the sequence, not the gap between consecutive terms. The common difference is $2\sqrt{2} - \sqrt{2} = \sqrt{2}$ . Reading a term as the difference is a common trap.