Question:easy

Which of the following sequence is not an A.P. ?

Show Hint

A sequence of squares of consecutive odd integers (or consecutive natural numbers) never forms an A.P. because the gap between consecutive squares increases quadratically.
This allows you to quickly identify option (D) as the correct choice without doing any calculations!
Updated On: Jul 7, 2026
  • \(2, \frac{5}{2}, 3, \frac{7}{2}, \dots\)
  • \(-1.2, -3.2, -5.2, -7.2, \dots\)
  • \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \dots\)
  • \(1^2, 3^2, 5^2, 7^2, \dots\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Use the general-term test instead of checking differences one by one.
A sequence is an A.P. exactly when its $n$th term can be written as a linear expression in $n$, that is $a_n = An + B$ for some constants $A$ and $B$. If the $n$th term instead depends on $n$ in a squared or otherwise non-linear way, the sequence cannot be an A.P.

Step 2: Check option (A).
The terms are $2, \frac{5}{2}, 3, \frac{7}{2}, \dots$, which increase by a fixed $\frac{1}{2}$ each time. Its $n$th term is $a_n = 2 + \frac{1}{2}(n-1)$, which is linear in $n$. This is an A.P.

Step 3: Check option (B).
The terms are $-1.2, -3.2, -5.2, -7.2, \dots$, decreasing by a fixed $2.0$ each time. The $n$th term is $a_n = -1.2 - 2(n-1)$, again linear in $n$. This is an A.P.

Step 4: Check option (C).
Simplify the radicals first: $\sqrt{2}$, $\sqrt{8} = 2\sqrt{2}$, $\sqrt{18} = 3\sqrt{2}$. So the terms are $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, \dots$, whose $n$th term is $a_n = n\sqrt{2}$, which is linear in $n$. This is an A.P.

Step 5: Check option (D).
The terms are $1^2, 3^2, 5^2, 7^2, \dots$, that is, the square of the $n$th odd number:
\[ a_n = (2n - 1)^2 = 4n^2 - 4n + 1 \]
This expression has an $n^2$ term, so it is quadratic in $n$, not linear. A sequence whose general term is quadratic cannot have a constant common difference, so this is not an A.P.

Step 6: Final Answer.
Only the sequence $1^2, 3^2, 5^2, 7^2, \dots$ fails the linear general-term test, so option (D) is correct. \[ \boxed{1^2,\ 3^2,\ 5^2,\ 7^2,\ \dots} \]
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