Question:
If the mean of the first \(n\) odd numbers is \( \frac{n^2}{81} \), then \(n\) equals
If the mean of the first \(n\) odd numbers is \( \frac{n^2}{81} \), then \(n\) equals
Show Hint
Always remember:
\[
1+3+5+\cdots+(2n-1)=n^2
\]
This is one of the most frequently used formulas in aptitude exams.
Updated On: Apr 30, 2026
- \(9\)
- \(18\)
- \(27\)
- \(81\)
- \(52\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Concept:
The first \(n\) odd numbers are:
\[
1,\ 3,\ 5,\ 7,\ \dots,\ (2n-1)
\]
A very important result is:
\[
\text{Sum of first } n \text{ odd numbers} = n^2
\]
Therefore,
\[
\text{Mean}
=
\frac{n^2}{n}
=
n
\]
Step 1: Use the mean formula. Given: \[ \text{Mean} = \frac{n^2}{81} \] But we know: \[ \text{Mean of first } n \text{ odd numbers} = n \] Therefore, \[ n = \frac{n^2}{81} \]
Step 2: Solve the equation. \[ 81n = n^2 \] \[ n^2 - 81n = 0 \] \[ n(n-81)=0 \] So, \[ n = 0 \text{or} n = 81 \] Since number of terms cannot be zero, \[ \boxed{n = 81} \] Hence, correct option is: \[ \boxed{(D)\ 81} \]
Step 1: Use the mean formula. Given: \[ \text{Mean} = \frac{n^2}{81} \] But we know: \[ \text{Mean of first } n \text{ odd numbers} = n \] Therefore, \[ n = \frac{n^2}{81} \]
Step 2: Solve the equation. \[ 81n = n^2 \] \[ n^2 - 81n = 0 \] \[ n(n-81)=0 \] So, \[ n = 0 \text{or} n = 81 \] Since number of terms cannot be zero, \[ \boxed{n = 81} \] Hence, correct option is: \[ \boxed{(D)\ 81} \]
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