Let $a_1, a_2, a_3, \ldots$ be in harmonic progression with $a_1 = 5$ and $a_{20} = 25$. If $n$ is the least positive integer for which $a_n < 0$, then the value of $4n - 100$ is
Step-by-Step Solution
Key Concept: The core concept of transforming a Harmonic Progression (H.P.) into an Arithmetic Progression (A.P.) by taking reciprocals ($b_n = 1/a_n$), and subsequently applying the standard formula for the $n$-th term of an A.P. ($b_n = b_1 + (n-1)d$). Furthermore, understanding the critical equivalence that for non-zero an , the condition $a_n < 0$ is exactly equivalent to $1/a_n < 0$ is essential for setting up the correct inequality to determine $n$.
$a_1, a_2, a_3, \ldots$ are in H.P.
$\Rightarrow \frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots$ are in A.P.
Let the common difference of A.P. be $d$.
\begin{align*}
\frac{1}{a_{20}} &= \frac{1}{a_1} + (20-1)d \\
\Rightarrow \frac{1}{a_{20}} - \frac{1}{a_1} &= 19d \\
\Rightarrow \frac{1}{25} - \frac{1}{5} &= 19d \\
\Rightarrow \frac{1-5}{25} &= 19d \\
\Rightarrow -\frac{4}{25} &= 19d \\
\Rightarrow d &= \frac{-4}{19 \times 25}
\end{align*}
We need to find the least positive integer $n$ such that $a_n < 0$, which implies $\frac{1}{a_n} < 0$.
\begin{align*}
\frac{1}{a_n} &= \frac{1}{a_1} + (n-1)d \\
\Rightarrow \frac{1}{a_n} &= \frac{1}{5} + (n-1)\left(\frac{-4}{19 \times 25}\right) \\
\text{So, we require } \frac{1}{5} + (n-1)\left(\frac{-4}{19 \times 25}\right) &< 0 \\
\Rightarrow \frac{1}{5} &< \frac{4(n-1)}{19 \times 25} \\
\Rightarrow \frac{4(n-1)}{19 \times 5} &> 1 \\
\Rightarrow n-1 &> \frac{19 \times 5}{4} \\
\Rightarrow n &> \frac{19 \times 5}{4} + 1 \\
\Rightarrow n &> \frac{95}{4} + 1 \\
\Rightarrow n &> 23.75 + 1 \\
\Rightarrow n &> 24.75 \\
\Rightarrow n &\geq 25 \quad \text{ (since } n \text{ is an integer)}
\end{align*}
The least positive integer value of $n$ is $25$.
Hence, the value of $4n - 100 = 4 \times 25 - 100 = 0$.
<div class="key-concept"><strong>Key Concept:</strong> The core concept of transforming a Harmonic Progression (H.P.) into an Arithmetic Progression (A.P.) by taking reciprocals ($b_n = 1/a_n$), and subsequently applying the standard formula for the $n$-th term of an A.P. ($b_n = b_1 + (n-1)d$). Furthermore, understanding the critical equivalence that for non-zero an , the condition $a_n < 0$ is exactly equivalent to $1/a_n < 0$ is essential for setting up the correct inequality to determine $n$.</div>
<div class="trap-box"><strong>Trap:</strong> The most common trap is incorrectly determining the least positive integer $n$ from the final inequality. After correctly deriving $n > 24.75$, students might mistakenly round down to $24$ or make an off-by-one error, instead of correctly identifying $n=25$ as the smallest integer strictly greater than $24.75$.</div>
Correct Answer: 0