Question

If the arithmetic mean of the numbers \[ 2,4,6,\ldots,2n \] is \(\frac{7}{4}\), then \(n\) is equal to:

• A. \(1\)
• B. \(\frac34\)
• C. \(\frac54\)
• D. \(\frac74\)

Hint:

For any arithmetic progression, \[ \text{Mean} = \frac{\text{First Term}+\text{Last Term}}{2}. \] This shortcut avoids lengthy summation calculations.

Answer 1

The Correct Option is B

Step 1: Understand the numbers.
The numbers are $2,4,6,\ldots,2n$. Each term is $2$ more than the one before, so this is an arithmetic progression with first term $2$ and common difference $2$.

Step 2: Recall the average of an A.P.
For an arithmetic progression, the average of all the terms equals the average of the first term and the last term. So the arithmetic mean is $\dfrac{\text{first}+\text{last}}{2}$.

Step 3: Plug in the first and last terms.
Here the first term is $2$ and the last term is $2n$. So the mean is \[ \frac{2+2n}{2}=1+n. \]

Step 4: Set the mean equal to the given value.
We are told this mean is $\dfrac{7}{4}$. So \[ 1+n=\frac{7}{4}. \]

Step 5: Solve for $n$.
Subtract $1$ from both sides. \[ n=\frac{7}{4}-1=\frac{7-4}{4}=\frac{3}{4}. \]

Step 6: State the result.
The value that satisfies the equation is $\dfrac{3}{4}$. \[ \boxed{\dfrac{3}{4}} \]