Question

If 2k, 3k-1, 8 are in A.P. Then the value of k is :

• A. $\frac{3}{2}$
• B. $\frac{5}{2}$
• C. $\frac{1}{2}$
• D. $\frac{7}{2}$

Hint:

For three terms $a, b, c$ to be in Arithmetic Progression (A.P.), the middle term must be the average of the other two, i.e., $2b = a + c$, or equivalently, the common difference between consecutive terms must be equal: $b-a = c-b$.

Answer 1

The Correct Option is B

Step 1: Understand the property of an Arithmetic Progression (A.P.).
In an Arithmetic Progression, the difference between consecutive terms is constant. This constant difference is called the common difference. If $a$, $b$, and $c$ are in A.P., then $b - a = c - b$. This can be rearranged to $2b = a + c$. Step 2: Apply the A.P. property to the given terms.
The given terms are $2k$, $3k-1$, and $8$.
Let $a = 2k$
Let $b = 3k-1$
Let $c = 8$
Using the property $2b = a + c$:
$2(3k-1) = 2k + 8$ Step 3: Solve the equation for k.
Distribute the 2 on the left side:
$6k - 2 = 2k + 8$
Subtract $2k$ from both sides:
$6k - 2k - 2 = 8$
$4k - 2 = 8$
Add 2 to both sides:
$4k = 8 + 2$
$4k = 10$
Divide by 4:
$k = \frac{10}{4}$
$k = \frac{5}{2}$
Step 4: Verify the result.
If $k = 5/2$:
The terms are:
$2k = 2(5/2) = 5$
$3k-1 = 3(5/2) - 1 = 15/2 - 2/2 = 13/2 = 6.5$
$8$
The sequence is $5, 6.5, 8$.
The common difference is $6.5 - 5 = 1.5$ and $8 - 6.5 = 1.5$.
Since the common difference is constant, the terms are in A.P. Step 5: Compare the result with the given options.
The calculated value of $k$ is $\frac{5}{2}$, which matches option (2). \[ (2) \quad \mathbf{\frac{5}{2}} \]