Which of the following sequence is **not** an A.P. ?
A. $2,\ \dfrac52,\ 3,\ \dfrac72,\ \dots$
B. $-1{\cdot}2,\ -3{\cdot}2,\ -5{\cdot}2,\ -7{\cdot}2,\ \dots$
C. $\sqrt2,\ \sqrt8,\ \sqrt{18},\ \dots$
D. $1^2,\ 3^2,\ 5^2,\ 7^2,\ \dots$
✅ Answer & Solution
✅ Correct Answer: D
**Step 1 — Recall the test.**
A sequence is an A.P. if the common difference $d = a_{n+1} - a_n$ is the **same** throughout.
**Step 2 — Check each option.**
**(A)** $2,\ \dfrac52,\ 3,\ \dfrac72,\dots$ → $d = \dfrac52 - 2 = \dfrac12$, $3 - \dfrac52 = \dfrac12$ ✔ A.P.
**(B)** $-1{\cdot}2,\ -3{\cdot}2,\ -5{\cdot}2,\ -7{\cdot}2,\dots$ → $d = -2$ throughout ✔ A.P.
**(C)** $\sqrt2,\ \sqrt8,\ \sqrt{18},\dots = \sqrt2,\ 2\sqrt2,\ 3\sqrt2,\dots$ → $d = \sqrt2$ ✔ A.P.
**(D)** $1^2,\ 3^2,\ 5^2,\ 7^2,\dots = 1,\ 9,\ 25,\ 49,\dots$
$$9-1 = 8,\quad 25-9 = 16,\quad 49-25 = 24$$
Differences are $8, 16, 24$ — **not equal**. ✘ Not an A.P.
**Answer : (D) $1^2, 3^2, 5^2, 7^2, \dots$**
✅ Verified by Super Admin