Question

Which of the following forms an Arithmetic Progression?

• A. \(1,\ \frac12,\ \frac14,\ \frac18,\dots\)
• B. \(\frac12,\ 1,\ \frac32,\ 2,\dots\)
• C. \(2,\ 5,\ 10,\ 17,\dots\)
• D. \(1,\ 2,\ 6,\ 24,\dots\)

Hint:

To quickly check whether a sequence is an A.P.: \[ \text{Subtract consecutive terms} \] If all differences are equal, it is an Arithmetic Progression. If ratios are equal instead of differences, then it is a Geometric Progression (G.P.).

Answer 1

The Correct Option is B

Concept: An Arithmetic Progression (A.P.) is a sequence in which the difference between consecutive terms remains constant. This constant value is called the common difference. If: \[ a_1,\ a_2,\ a_3,\dots \] is an A.P., then: \[ a_2-a_1 = a_3-a_2 = a_4-a_3 \]

Step 1: Check Option (1). Sequence: \[ 1,\ \frac12,\ \frac14,\ \frac18,\dots \] Find consecutive differences: \[ \frac12-1=-\frac12 \] \[ \frac14-\frac12=-\frac14 \] \[ \frac18-\frac14=-\frac18 \] The differences are not equal. Therefore, Option (1) is not an A.P.

Step 2: Check Option (2). Sequence: \[ \frac12,\ 1,\ \frac32,\ 2,\dots \] Find consecutive differences: \[ 1-\frac12=\frac12 \] \[ \frac32-1=\frac12 \] \[ 2-\frac32=\frac12 \] All differences are equal. Therefore, this sequence forms an Arithmetic Progression. Common difference: \[ d=\frac12 \]

Step 3: Check Option (3). Sequence: \[ 2,\ 5,\ 10,\ 17,\dots \] Differences: \[ 5-2=3 \] \[ 10-5=5 \] \[ 17-10=7 \] Differences are not equal. Therefore, Option (3) is not an A.P.

Step 4: Check Option (4). Sequence: \[ 1,\ 2,\ 6,\ 24,\dots \] Differences: \[ 2-1=1 \] \[ 6-2=4 \] \[ 24-6=18 \] Differences are not constant. Therefore, Option (4) is not an A.P.

Step 5: Conclude the correct option. Only Option (2) has a constant common difference. Hence it is an Arithmetic Progression. Final Answer: \[ \boxed{\frac12,\ 1,\ \frac32,\ 2,\dots} \]