If the first and the (2n-1)st terms of an A.P., a G.P. and an H.P. are equal and their nth terms are

Q: Wait, why is the $n^{th}$ term the middle term? Shouldn't it be something else?

Think about the total number of terms, which is $2n-1$. This is an odd number. In any sequence with an odd number of terms, there is exactly one middle term. To find its position, we calculate the average of the first and last indices: $\frac{1 + (2n - 1)}{2} = \frac{2n}{2} = n$. So yes, the $n^{th}$ term is perfectly centered!

Q: Why can we assume the terms $x$ and $y$ are positive?

In the context of Geometric and Harmonic Progressions, the definitions of Geometric Mean ($\sqrt{xy}$) and Harmonic Mean ($\frac{2xy}{x+y}$) are standardly defined for positive real numbers. If they were negative, the square root would be imaginary, which doesn't fit the standard context of these progressions in JEE problems.

Q: Is $a=b=c$ always true?

Not necessarily! The inequality $a \ge b \ge c$ holds generally. However, if the first term $x$ and the last term $y$ are equal, the entire sequence becomes constant. In that specific case, the Arithmetic, Geometric, and Harmonic means all collapse to the same value, making $a = b = c$.

Q: Why is $ac - b^2 = 0$ a correct option?

This comes from the beautiful identity $G^2 = AH$. Since we identified $a=A, b=G,$ and $c=H$, substituting these gives $b^2 = ac$. Rearranging this equation leads directly to $ac - b^2 = 0$.

Question

If the first and the $(2n-1)$st terms of an A.P., a G.P. and an H.P. are equal and their $n$th terms are $a, b$ and $c$ respectively, then

Options:

(A) $a = b = c$

(B) $a \ge b \ge c$

(C) $a + c = b$

(D) $ac - b^2 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the starting line of a grand mathematical race track. You have three runners: an Arithmetic Progression (A.P.), a Geometric Progression (G.P.), and a Harmonic Progression (H.P.).

They all start at the same point, which we will call $x$, and they are all destined to finish at the same point, $y$. The race consists of exactly $2n-1$ steps.

### The Revelation of the Middle Term

Before we analyze how these runners move, we must identify the heart of the race. With $2n-1$ steps, there is a single, solitary middle step.

To find it, we calculate the index of the middle term:

$$n = \frac{1 + (2n - 1)}{2}$$

This means the $n^{th}$ term is the pivot point of our entire sequence and the point of symmetry. The problem states that at this $n^{th}$ step, our runners are at positions $a$, $b$, and $c$ respectively.

### The Three Pillars of Means

Now, let us look at how each runner behaves. The A.P. runner moves with a constant difference, meaning the middle term is the Arithmetic Mean of its extremes:

$$a = \frac{x+y}{2}$$

The G.P. runner moves with a constant ratio, meaning the middle term is the Geometric Mean of its extremes:

$$b = \sqrt{xy}$$

Finally, the H.P. runner has a middle term that is the Harmonic Mean of its extremes:

$$c = \frac{2xy}{x+y}$$

We have mapped the abstract variables $a$, $b$, and $c$ to the three most iconic quantities in algebra: the A.M., G.M., and H.M.

### The Grand Unification

There is a universal law in mathematics, the $A \ge G \ge H$ inequality. For any two positive numbers, the Arithmetic Mean is always greater than or equal to the Geometric Mean, which in turn is greater than or equal to the Harmonic Mean.

By substituting our variables, we immediately see that:

$$a \ge b \ge c$$

There is a deeper, more elegant connection between these means. The square of the Geometric Mean is always equal to the product of the Arithmetic and Harmonic Means:

$$b^2 = ac \quad \Rightarrow \quad ac - b^2 = 0$$

This is a fundamental identity that binds these progressions together.

### Conclusion: The Beauty of Symmetry

We started with three different runners on a track, and through the lens of the $n^{th}$ term, we discovered that they are all dancing to the same tune. Whether it is the inequality $a \ge b \ge c$ or the identity **$ac - b^2 = 0$**, we have uncovered the hidden structure of these sequences.

Remember, in JEE Advanced, it is rarely about brute force calculation; it is about recognizing the underlying geometric and algebraic soul of the problem. You have mastered the means; now go forth and master the exam.

Animated Solution for Mathematics - Sequence and Series: If the first and the $(2 n - 1)$ st terms of an A.P., a G.P. and an H.P. are equal and their $n$ th terms are $a, b$ and $c$ respectively, then

Select Answer:

Visualized Solution

Define Boundary Terms

Identify the Middle Term

A.P. and Arithmetic Mean

G.P. and Geometric Mean

H.P. and Harmonic Mean

The $A \geq G \geq H$ Inequality

Equality Condition

The $G^{2} = A H$ Relation

Conclusion and Summary

The Sigma Insight: Relation Between A.M., G.M., and H.M.

Analyzing the Setup

The Revelation of the Middle Term

The Three Pillars of Means

The Grand Unification

Conclusion

Similar Questions

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Select Answer:

Visualized Solution

Define Boundary Terms

Identify the Middle Term

A.P. and Arithmetic Mean

G.P. and Geometric Mean

H.P. and Harmonic Mean

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A≥G≥H Inequality

Equality Condition

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G2=AH Relation

Conclusion and Summary

The Sigma Insight: Relation Between A.M., G.M., and H.M.

Analyzing the Setup

The Revelation of the Middle Term

The Three Pillars of Means

The Grand Unification

Conclusion

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