If a1, a2, ..., an are in H.P., then the expression a1a2 + a2a3 + ... + a_{n-1}an is equal to (A) n(

Q: Why do we always convert H.P. to A.P.? Can't we just work with the H.P. terms directly?

That is a fantastic question! The reason is structural. Harmonic Progressions are defined by their reciprocals. While A.P.s have a linear, additive structure that is easy to manipulate, H.P.s are inherently non-linear. By converting to A.P., we move from a world of complex fractions to the elegant, predictable world of linear arithmetic. It is not just a trick; it is the fundamental language of the sequence.

Q: I am looking at the telescoping sum, and I am confused. Why do the terms cancel out? Is it always guaranteed?

It is guaranteed as long as the structure is consistent! A telescoping sum works like a row of falling dominoes. When you write the sum as $(a_1 - a_2) + (a_2 - a_3) + (a_3 - a_4) \dots$, the $-a_2$ from the first term is immediately neutralized by the $+a_2$ from the second term. This happens for every internal term. It is a beautiful, self-correcting mechanism that leaves only the boundary conditions—the start and the end—intact.

Q: Is there a faster way to find the common difference $d$ without using the $n^{th}$ term formula?

While you could potentially find $d$ through other algebraic manipulations, the $n^{th}$ term formula $\frac{1}{a_n} = \frac{1}{a_1} + (n-1)d$ is the most robust and reliable method. It connects the start and end of your sequence directly. In JEE Advanced, reliability is just as important as speed. Stick to the standard definition; it prevents errors and builds a solid foundation for more complex problems.

Question

If $a_1, a_2, \dots, a_n$ are in H.P., then the expression $a_1 a_2 + a_2 a_3 + \dots + a_{n-1} a_n$ is equal to

Options:

(A) $n(a_1 - a_n)$

(B) $(n-1)(a_1 - a_n)$

(C) $n a_1 a_n$

(D) $(n-1) a_1 a_n$

Sigma Percentile · Accepted Answer

### The Transformation: From Chaos to Order

When we are given a sequence $a_1, a_2, \dots, a_n$ in Harmonic Progression (H.P.), our first instinct might be to panic because H.P. does not have a simple additive property. However, the definition of H.P. is a gift.

It tells us that the reciprocals of these terms, $\frac{1}{a_1}, \frac{1}{a_2}, \dots, \frac{1}{a_n}$, form a perfect, orderly Arithmetic Progression (A.P.).

Let us visualize this on a number line. Imagine these reciprocals as points spaced at perfectly equal intervals. The distance between any two consecutive points is a constant, which we call the common difference, $d$. Mathematically, we write this as:

$$\frac{1}{a_{k+1}} - \frac{1}{a_k} = d$$

### The Algebraic Magic: Turning Products into Differences

Now, look at the expression we need to evaluate: $S = a_1 a_2 + a_2 a_3 + \dots + a_{n-1} a_n$. This is a sum of products of consecutive terms.

Using our A.P. definition, we take the equation $\frac{1}{a_{k+1}} - \frac{1}{a_k} = d$. If we find a common denominator, we get:

$$\frac{a_k - a_{k+1}}{a_k a_{k+1}} = d$$

Now, watch closely as we rearrange this to isolate the product $a_k a_{k+1}$:

$$a_k a_{k+1} = \frac{a_k - a_{k+1}}{d}$$

We have transformed a product of two terms into a difference of two terms, all divided by a constant $d$. This is the turning point of the problem.

### The Telescoping Collapse

Now, let us substitute this identity back into our sum $S$. We get:

$$S = \frac{1}{d} [ (a_1 - a_2) + (a_2 - a_3) + \dots + (a_{n-1} - a_n) ]$$

Look at the terms inside the bracket. We have $-a_2$ followed by $+a_2$, $-a_3$ followed by $+a_3$, and so on. This is what we call a **telescoping sum**.

It is like a chain reaction where every intermediate term cancels out, leaving only the first and the last. The sum collapses, and we are left with:

$$S = \frac{a_1 - a_n}{d}$$

### Closing the Loop

We are almost there! We have the expression in terms of $a_1$, $a_n$, and $d$, but we need to eliminate $d$. We go back to our A.P. definition where the $n^{th}$ term is given by:

$$\frac{1}{a_n} = \frac{1}{a_1} + (n-1)d$$

Rearranging this to solve for $(n-1)d$, we get:

$$(n-1)d = \frac{1}{a_n} - \frac{1}{a_1} = \frac{a_1 - a_n}{a_1 a_n}$$

This gives us the value of $d$:

$$d = \frac{a_1 - a_n}{(n-1) a_1 a_n}$$

Now, substitute this back into our simplified sum $S = \frac{a_1 - a_n}{d}$. When you divide by $d$, you are multiplying by its reciprocal. The term $(a_1 - a_n)$ in the numerator and the denominator cancels out perfectly, leaving us with the elegant result:

**$$S = (n-1) a_1 a_n$$**

Take a moment to appreciate this. We started with a complex, intimidating series, and through the power of transformation and telescoping, we reduced it to a simple, clean expression.

Animated Solution for Mathematics - Sequence and Series: If $a_{1}, a_{2}, \dots, a_{n}$ are in H.P., then the expression $a_{1} a_{2} + a_{2} a_{3} + \dots + a_{n - 1} a_{n}$ is equal to

Select Answer:

Visualized Solution

Understanding Harmonic Progression

Defining the Common Difference $d$

Expressing the Product $a_{k} a_{k + 1}$

Expanding the Required Sum

The Telescoping Cancellation

Finding the Total Distance $(n - 1) d$

Expressing $d$ in terms of $a_{1}$ and $a_{n}$

Final Substitution & Simplification

Key Takeaways

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

The Transformation

The Algebraic Magic

The Telescoping Collapse

Closing the Loop

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Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a1​,a2​,…,an​ are in H.P., then the expression a1​a2​+a2​a3​+⋯+an−1​an​ is equal to