Let a1, a2, a3 ... be in harmonic progression with a1 = 5 and a20 = 25. The least positive integer n

Q: Wait, why do we flip the terms? Why can't we just work with the H.P. directly?

That is a fantastic question! Harmonic Progressions are defined by the property that their reciprocals form an Arithmetic Progression. There is no simple, linear formula for the $n^{th}$ term of an H.P. like there is for an A.P. or G.P. By taking the reciprocal, we transform a complex, non-linear relationship into a simple, linear one. It is like changing your coordinate system to make the math work for you, not against you.

Q: Why does $a_n < 0$ imply $A_n < 0$? Is it always true?

Yes, it is! Think about the sign of a fraction $\frac{1}{x}$. If $x$ is positive, $\frac{1}{x}$ is positive. If $x$ is negative, $\frac{1}{x}$ is also negative. The reciprocal operation preserves the sign of the number. Since we are looking for when the H.P. term $a_n$ becomes negative, we are effectively looking for when its reciprocal $A_n$ crosses the x-axis into the negative region.

Q: I got $n > 24.75$. Why is the answer 25 and not 24?

Remember, $n$ represents the position of a term in a sequence, so it must be a positive integer. If $n$ must be strictly greater than $24.75$, then $24$ is not a solution because $24$ is less than $24.75$. The first integer that satisfies the condition $n > 24.75$ is $25$. Always check if your inequality requires the next integer up or down!

Question

Let $a_1, a_2, a_3 \dots$ be in harmonic progression with $a_1 = 5$ and $a_{20} = 25$. The least positive integer $n$ for which $a_n < 0$ is

Options:

(A) 22

(B) 23

(C) 24

(D) 25

Sigma Percentile · Accepted Answer

### The Hidden Symmetry of Harmonic Progressions Have you ever felt that some mathematical sequences are just... elusive? Harmonic Progressions (H.P.) often feel that way. They don't behave like the predictable Arithmetic Progressions (A.P.) or the explosive Geometric Progressions (G.P.). But here is the secret: H.P. is just an A.P. in disguise. To solve this problem, we don't need to fight the H.P.; we just need to strip away its mask. ### Phase 1: The Reciprocal Bridge We are given a sequence $a_1, a_2, a_3, \dots$ in Harmonic Progression. The definition of an H.P. is that the reciprocals of its terms form an Arithmetic Progression. Let us define a new sequence $A_n = \frac{1}{a_n}$. This new sequence $A_n$ is an A.P. with a common difference $d$. By shifting our focus to $A_n$, we turn a problem that seemed impossible into a standard linear equation problem. Imagine you are standing on a wedge, looking at the slope of this A.P.; it is a straight line, and we just need to find its equation. ### Phase 2: Mapping the Coordinates We are given two vital anchor points: $a_1 = 5$ and $a_{20} = 25$. Using our reciprocal bridge, we immediately find the corresponding points for our A.P.: $$A_1 = \frac{1}{a_1} = \frac{1}{5}$$ $$A_{20} = \frac{1}{a_{20}} = \frac{1}{25}$$ We now have two points on a line: $(1, \frac{1}{5})$ and $(20, \frac{1}{25})$. The general term of an A.P. is given by $A_n = A_1 + (n-1)d$. For the 20th term, this becomes $A_{20} = A_1 + 19d$. ### Phase 3: The Descent Now, let us find the common difference $d$. Substituting our known values into the equation: $$\frac{1}{25} = \frac{1}{5} + 19d$$ To isolate $d$, we subtract $\frac{1}{5}$ from both sides. Remember, $\frac{1}{5}$ is the same as $\frac{5}{25}$. So, we have: $$19d = \frac{1}{25} - \frac{5}{25} = -\frac{4}{25}$$ Dividing by $19$, we get the common difference: $$d = -\frac{4}{25 imes 19} = -\frac{4}{475}$$ The negative sign is beautiful—it tells us that our A.P. is decreasing. As $n$ increases, the terms of the A.P. are getting smaller, which means the terms of the H.P. are getting closer to zero and eventually crossing into the negative territory. ### Phase 4: The Inequality Threshold We want to find the least positive integer $n$ such that $a_n < 0$. As we discussed, this is equivalent to $A_n < 0$. Let us set up the inequality: $$A_1 + (n-1)d < 0$$ $$\frac{1}{5} + (n-1)\left(-\frac{4}{475} ight) < 0$$ To make this easier, let us multiply the entire inequality by $475$ to clear the denominators: $$475 imes \frac{1}{5} - 4(n-1) < 0$$ $$95 - 4(n-1) < 0$$ Now, expand the expression: $$95 - 4n + 4 < 0$$ $$99 - 4n < 0$$ Rearranging to solve for $n$: $$4n > 99$$ $$n > \frac{99}{4}$$ $$n > 24.75$$ ### Conclusion: The Final Step We have arrived at $n > 24.75$. Since $n$ must be a positive integer, the smallest integer that satisfies this condition is **$25$**. It is a moment of triumph! We started with a complex Harmonic Progression, transformed it into a simple Arithmetic Progression, navigated the algebra, and found the exact point where the sequence dips below zero. Mathematics is not about memorizing formulas; it is about finding the right perspective to see the underlying structure. You have mastered this!

Animated Solution for Mathematics - Sequence and Series: Let $a_{1}, a_{2}, a_{3} \dots$ be in harmonic progression with $a_{1} = 5$ and $a_{20} = 25$ . The least positive integer $n$ for which $a_{n} < 0$ is

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Visualized Solution

Harmonic to Arithmetic Progression

Identifying Given Values

The 20th Term

Applying the A.P. Formula

Substituting the Values

Solving for Common Difference $d$

Setting the Condition $a_{n} < 0$

Formulating the Inequality

Simplifying the Inequality

Expanding the Expression

Solving for $n$

Finding the Least Integer

The Sigma Insight: Harmonic Progression (H.P.)

The Hidden Symmetry of Harmonic Progressions

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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