Question:
If the arithmetic mean of the numbers
\[
2,4,6,\ldots,2n
\]
is \(\frac{7}{4}\), then \(n\) is equal to:
If the arithmetic mean of the numbers
\[
2,4,6,\ldots,2n
\]
is \(\frac{7}{4}\), then \(n\) is equal to:
Show Hint
For any arithmetic progression,
\[
\text{Mean}
=
\frac{\text{First Term}+\text{Last Term}}{2}.
\]
This shortcut avoids lengthy summation calculations.
Updated On: Jun 10, 2026
- \(1\)
- \(\frac34\)
- \(\frac54\)
- \(\frac74\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
The sequence
\[
2,4,6,\ldots,2n
\]
forms an arithmetic progression because the difference between consecutive terms remains constant.
For any arithmetic progression, the arithmetic mean of all terms is equal to the average of the first and the last term.
That is,
\[
\text{A.M.}
=
\frac{\text{First Term}+\text{Last Term}}{2}.
\]
This property eliminates the need to calculate the entire sum of the sequence.
Step 1: Identify the first and last terms. The first term is \[ 2. \] The last term is \[ 2n. \]
Step 2: Write the formula for arithmetic mean. Using \[ \text{A.M.} = \frac{\text{First Term}+\text{Last Term}}{2}, \] we get \[ \text{A.M.} = \frac{2+2n}{2}. \]
Step 3: Simplify the expression. \[ \text{A.M.} = \frac{2(1+n)}{2}. \] \[ =n+1. \]
Step 4: Use the given value of the mean. Given that \[ n+1 = \frac74. \] Subtracting \(1\), \[ n = \frac74-\frac44. \] \[ = \frac34. \]
Step 5: Verification. Substituting \[ n=\frac34, \] into \[ n+1, \] gives \[ \frac34+1 = \frac74. \] This agrees with the condition.
Step 6: Final Conclusion. \[ \boxed{n=\frac34} \] Hence the correct answer is \[ \boxed{\text{Option (B)}}. \]
Step 1: Identify the first and last terms. The first term is \[ 2. \] The last term is \[ 2n. \]
Step 2: Write the formula for arithmetic mean. Using \[ \text{A.M.} = \frac{\text{First Term}+\text{Last Term}}{2}, \] we get \[ \text{A.M.} = \frac{2+2n}{2}. \]
Step 3: Simplify the expression. \[ \text{A.M.} = \frac{2(1+n)}{2}. \] \[ =n+1. \]
Step 4: Use the given value of the mean. Given that \[ n+1 = \frac74. \] Subtracting \(1\), \[ n = \frac74-\frac44. \] \[ = \frac34. \]
Step 5: Verification. Substituting \[ n=\frac34, \] into \[ n+1, \] gives \[ \frac34+1 = \frac74. \] This agrees with the condition.
Step 6: Final Conclusion. \[ \boxed{n=\frac34} \] Hence the correct answer is \[ \boxed{\text{Option (B)}}. \]
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