Question:
The first term of an A.P. is $148$ and the common difference is $-2$. If the A.M. of first $n$ terms of the A.P. is $125$, then the value of $n$ is
The first term of an A.P. is $148$ and the common difference is $-2$. If the A.M. of first $n$ terms of the A.P. is $125$, then the value of $n$ is
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Mean of A.P. simplifies nicely to $\frac{1}{2}[2a+(n-1)d]$, no need to compute full sum.
Updated On: May 8, 2026
- $18$
- $24$
- $30$
- $36$
- $48$
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The Correct Option is B
Solution and Explanation
Concept:
• Sum of A.P.: $\displaystyle S_n = \frac{n}{2}[2a + (n-1)d]$
• Mean of first $n$ terms: \[ \text{A.M.} = \frac{S_n}{n} \]
Step 1: Write expression for mean
\[ \frac{S_n}{n} = 125 \] \[ \frac{1}{n} \cdot \frac{n}{2}[2a + (n-1)d] = 125 \] \[ \frac{1}{2}[2a + (n-1)d] = 125 \]
Step 2: Substitute values
\[ a = 148,\quad d = -2 \] \[ \frac{1}{2}[2(148) + (n-1)(-2)] = 125 \] \[ \frac{1}{2}[296 -2(n-1)] = 125 \]
Step 3: Simplify
\[ \frac{1}{2}[296 -2n +2] = 125 \] \[ \frac{1}{2}[298 -2n] = 125 \] \[ 149 - n = 125 \]
Step 4: Solve
\[ n = 149 -125 = 24 \] \[ \boxed{24} \]
• Sum of A.P.: $\displaystyle S_n = \frac{n}{2}[2a + (n-1)d]$
• Mean of first $n$ terms: \[ \text{A.M.} = \frac{S_n}{n} \]
Step 1: Write expression for mean
\[ \frac{S_n}{n} = 125 \] \[ \frac{1}{n} \cdot \frac{n}{2}[2a + (n-1)d] = 125 \] \[ \frac{1}{2}[2a + (n-1)d] = 125 \]
Step 2: Substitute values
\[ a = 148,\quad d = -2 \] \[ \frac{1}{2}[2(148) + (n-1)(-2)] = 125 \] \[ \frac{1}{2}[296 -2(n-1)] = 125 \]
Step 3: Simplify
\[ \frac{1}{2}[296 -2n +2] = 125 \] \[ \frac{1}{2}[298 -2n] = 125 \] \[ 149 - n = 125 \]
Step 4: Solve
\[ n = 149 -125 = 24 \] \[ \boxed{24} \]
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