Key Concepts and Formulas

• Harmonic Progression (H.P.): A sequence ${a_1, a_2, ..., a_n}$ is in H.P. if their reciprocals ${1/a_1, 1/a_2, ..., 1/a_n}$ are in Arithmetic Progression (A.P.).
• Arithmetic Progression (A.P.): A sequence ${b_1, b_2, ..., b_n}$ is in A.P. if the difference between consecutive terms is constant, i.e., ${b_{k+1} - b_k = d}$ for all ${k}$. The general term of an A.P. is ${b_k = b_1 + (k-1)d}$.
• Summation Notation: The expression ${a_1\,{a_2} + \,{a_2}\,{a_3}\, + .... + {a_{n - 1}}\,{a_n}}$ can be written in summation notation as ${{\sum_{i=1}^{n-1} a_i a_{i+1}}}$

Step-by-Step Solution

Step 1: Convert the H.P. to an A.P. Given that ${a_1, a_2, ..., a_n}$ are in H.P. By definition, their reciprocals ${1/a_1, 1/a_2, ..., 1/a_n}$ are in A.P. Let's denote these reciprocals as ${b_1, b_2, ..., b_n}$, where ${b_i = 1/a_i}$.

Step 2: Express the terms of the A.P. in relation to each other. Since ${b_1, b_2, ..., b_n}$ form an A.P., let ${d}$ be the common difference. Then, ${b_{i+1} - b_i = d}$ for all ${i = 1, 2, ..., n-1}$. This implies ${b_{i+1} = b_i + d}$.

Step 3: Express the product of consecutive terms of the H.P. in terms of A.P. terms. We need to evaluate the expression ${S = a_1\,a_2 + a_2\,a_3 + ... + a_{n-1}\,a_n}$. We can rewrite each product term ${a_i a_{i+1}}$ using the relationship ${a_i = 1/b_i}$. So, ${a_i a_{i+1} = (1/b_i) \times (1/b_{i+1}) = 1 / (b_i b_{i+1})}$. The expression becomes ${S = \frac{1}{b_1 b_2} + \frac{1}{b_2 b_3} + ... + \frac{1}{b_{n-1} b_n}}$.

Step 4: Manipulate the terms of the sum to utilize the A.P. property. Consider a generic term in the sum: ${1 / (b_i b_{i+1})}$. We know that ${b_{i+1} - b_i = d}$. Let's try to express ${1 / (b_i b_{i+1})}$ in a form that involves the difference ${d}$. Multiply the numerator and denominator by ${d}$: ${ \frac{1}{b_i b_{i+1}} = \frac{1}{d} \times \frac{d}{b_i b_{i+1}} }$ Now, substitute ${d = b_{i+1} - b_i}$ in the numerator: ${ \frac{1}{b_i b_{i+1}} = \frac{1}{d} \times \frac{b_{i+1} - b_i}{b_i b_{i+1}} }$ Separate the terms in the numerator: ${ \frac{1}{b_i b_{i+1}} = \frac{1}{d} \left( \frac{b_{i+1}}{b_i b_{i+1}} - \frac{b_i}{b_i b_{i+1}} \right) }$ This simplifies to: ${ \frac{1}{b_i b_{i+1}} = \frac{1}{d} \left( \frac{1}{b_i} - \frac{1}{b_{i+1}} \right) }$

Step 5: Apply the derived formula to each term in the summation. Now, we can substitute this back into our sum ${S = \sum_{i=1}^{n-1} \frac{1}{b_i b_{i+1}}}$. ${ S = \sum_{i=1}^{n-1} \frac{1}{d} \left( \frac{1}{b_i} - \frac{1}{b_{i+1}} \right) }$ Factor out the constant ${1/d}$: ${ S = \frac{1}{d} \sum_{i=1}^{n-1} \left( \frac{1}{b_i} - \frac{1}{b_{i+1}} \right) }$ This is a telescoping sum. Let's expand it: ${ S = \frac{1}{d} \left[ \left( \frac{1}{b_1} - \frac{1}{b_2} \right) + \left( \frac{1}{b_2} - \frac{1}{b_3} \right) + \left( \frac{1}{b_3} - \frac{1}{b_4} \right) + ... + \left( \frac{1}{b_{n-1}} - \frac{1}{b_n} \right) \right] }$ The intermediate terms cancel out: ${ S = \frac{1}{d} \left( \frac{1}{b_1} - \frac{1}{b_n} \right) }$

Step 6: Substitute back the original H.P. terms. Recall that ${b_i = 1/a_i}$. So, ${1/b_i = a_i}$. ${ S = \frac{1}{d} (a_1 - a_n) }$

Step 7: Determine the common difference ${d}$ in terms of ${a_1}$ and ${a_n}$. We know that ${b_1, b_2, ..., b_n}$ is an A.P. The first term is ${b_1 = 1/a_1}$. The last term is ${b_n = 1/a_n}$. In an A.P. with ${n}$ terms, the last term ${b_n}$ can be expressed as ${b_n = b_1 + (n-1)d}$. Substituting the values: ${ \frac{1}{a_n} = \frac{1}{a_1} + (n-1)d }$ Now, solve for ${d}$: ${ (n-1)d = \frac{1}{a_n} - \frac{1}{a_1} }$ ${ (n-1)d = \frac{a_1 - a_n}{a_1 a_n} }$ ${ d = \frac{a_1 - a_n}{(n-1)a_1 a_n} }$

Step 8: Substitute the expression for ${d}$ back into the expression for ${S}$. We have ${S = \frac{1}{d} (a_1 - a_n)}$. Substitute the value of ${d}$: ${ S = \frac{1}{\frac{a_1 - a_n}{(n-1)a_1 a_n}} (a_1 - a_n) }$ ${ S = \frac{(n-1)a_1 a_n}{a_1 - a_n} (a_1 - a_n) }$ Assuming ${a_1 - a_n \neq 0}$, we can cancel out the ${(a_1 - a_n)}$ term: ${ S = (n-1)a_1 a_n }$

Correction and Re-evaluation: Let's re-examine Step 5 and 6. The derivation ${S = \frac{1}{d} (a_1 - a_n)}$ is correct. The mistake might be in the interpretation or the final substitution. Let's go back to the expression for ${d}$. ${ d = \frac{a_1 - a_n}{(n-1)a_1 a_n} }$ Then ${ \frac{1}{d} = \frac{(n-1)a_1 a_n}{a_1 - a_n} }$ So, ${ S = \frac{(n-1)a_1 a_n}{a_1 - a_n} (a_1 - a_n) = (n-1)a_1 a_n }$. This result matches option (D), but the correct answer is (A). Let's check the problem statement and the options again.

There seems to be a misunderstanding in my derivation or approach. Let's reconsider the problem from the beginning, focusing on the relationship between ${a_i}$ and ${b_i}$.

Let ${b_i = 1/a_i}$. Then ${b_1, b_2, ..., b_n}$ is an A.P. with common difference ${d}$. So, ${b_{i+1} = b_i + d}$. We want to evaluate ${ \sum_{i=1}^{n-1} a_i a_{i+1} }$ ${ a_i a_{i+1} = \frac{1}{b_i b_{i+1}} }$ We used ${ \frac{1}{b_i b_{i+1}} = \frac{1}{d} \left( \frac{1}{b_i} - \frac{1}{b_{i+1}} \right) }$. This is correct. The sum is ${ S = \frac{1}{d} \left( \frac{1}{b_1} - \frac{1}{b_n} \right) }$ which is ${ S = \frac{1}{d} (a_1 - a_n) }$.

Let's re-examine how ${d}$ is related to the original terms. ${ b_n = b_1 + (n-1)d }$ ${ \frac{1}{a_n} = \frac{1}{a_1} + (n-1)d }$ ${ (n-1)d = \frac{1}{a_n} - \frac{1}{a_1} = \frac{a_1 - a_n}{a_1 a_n} }$

Consider the case when ${n=2}$. The sum is ${a_1 a_2}$. Option (A): ${2(a_1 - a_2)}$. Option (B): ${(2-1)(a_1 - a_2) = a_1 - a_2}$. Option (C): ${2 a_1 a_2}$. Option (D): ${(2-1) a_1 a_2 = a_1 a_2}$. For ${n=2}$, the expression is ${a_1 a_2}$. This matches option (D). However, the provided correct answer is (A). This indicates a potential issue with the problem statement, options, or the provided correct answer, or a fundamental misunderstanding on my part.

Let's assume the correct answer (A) is indeed correct and try to work backward or find a scenario where it holds. If the answer is ${n(a_1 - a_n)}$, then ${S = n(a_1 - a_n)}$.

Let's re-check the telescoping sum step. ${ S = \frac{1}{d} \sum_{i=1}^{n-1} \left( \frac{1}{b_i} - \frac{1}{b_{i+1}} \right) = \frac{1}{d} \left( \frac{1}{b_1} - \frac{1}{b_n} \right) }$ This is ${S = \frac{1}{d} (a_1 - a_n)}$.

Consider a specific H.P.: ${6, 3, 2}$. Here ${n=3}$. The reciprocals are ${1/6, 1/3, 1/2}$. This is an A.P. with first term ${1/6}$ and common difference ${1/3 - 1/6 = 1/6}$. So, ${b_1 = 1/6, b_2 = 1/3, b_3 = 1/2}$. ${a_1 = 6, a_2 = 3, a_3 = 2}$. The expression is ${a_1 a_2 + a_2 a_3 = 6 \times 3 + 3 \times 2 = 18 + 6 = 24}$.

Let's check the options for this case (${n=3, a_1=6, a_3=2}$): (A) ${n(a_1 - a_n) = 3(6 - 2) = 3 \times 4 = 12}$. This does not match 24. (B) ${(n - 1)(a_1 - a_n) = (3 - 1)(6 - 2) = 2 \times 4 = 8}$. This does not match 24. (C) ${n a_1 a_n = 3 \times 6 \times 2 = 36}$. This does not match 24. (D) ${(n - 1) a_1 a_n = (3 - 1) \times 6 \times 2 = 2 \times 12 = 24}$. This matches 24.

My derivation leads to option (D), and the example confirms it. However, the problem statement insists that the correct answer is (A). This implies there might be a typo in the question, the options, or the provided correct answer.

Let's assume, for the sake of arriving at the given correct answer (A), that there is a specific condition or a different interpretation.

If the question intended to ask for ${ \sum_{i=1}^{n-1} \frac{1}{a_i a_{i+1}} }$, then the result would be { \frac{1}{d} (a_1 - a_n) }$. If {d}$$ was related differently, or if the summation was different.

Let's reconsider the possibility of a typo in the question or options. If the question was about a different expression, or if the options were different.

Given the discrepancy, and assuming the provided correct answer (A) is indeed correct, it suggests that my standard approach to H.P. problems might be missing a nuance specific to this problem or that there is an error in the problem statement/options.

However, based on the standard definitions and manipulation of H.P. and A.P., the derivation ${ \sum_{i=1}^{n-1} a_i a_{i+1} = (n-1)a_1 a_n }$ appears to be correct, as validated by the example.

Let's consider if the question implicitly assumes something about the common difference of the A.P. formed by the reciprocals.

Let's assume there is a typo in the question and it should be ${ \sum_{i=1}^{n-1} (a_{i+1} - a_i) }$ or something similar. But the given expression is clear.

Given the provided correct answer is (A), and my derived answer is (D) which is also verified by an example, there is a strong indication of an error in the problem's stated correct answer.

However, if forced to choose an approach that might lead to (A), it would require a significantly different manipulation or a misunderstanding of the problem.

Let's assume there is a formula related to H.P. that I am not recalling or applying correctly.

Let's write down the relationships again: ${b_i = 1/a_i}$ ${b_{i+1} = b_i + d}$ ${b_n = b_1 + (n-1)d}$ ${d = \frac{b_n - b_1}{n-1} = \frac{1/a_n - 1/a_1}{n-1} = \frac{a_1 - a_n}{(n-1)a_1 a_n}}$ The sum is ${S = \sum_{i=1}^{n-1} a_i a_{i+1} = \sum_{i=1}^{n-1} \frac{1}{b_i b_{i+1}} }$ ${ \frac{1}{b_i b_{i+1}} = \frac{1}{d} \left( \frac{1}{b_i} - \frac{1}{b_{i+1}} \right) }$ ${ S = \frac{1}{d} \left( \frac{1}{b_1} - \frac{1}{b_n} \right) = \frac{1}{d} (a_1 - a_n) }$ Substituting ${d = \frac{a_1 - a_n}{(n-1)a_1 a_n}}$: ${ S = \frac{(n-1)a_1 a_n}{a_1 - a_n} (a_1 - a_n) = (n-1)a_1 a_n }$

If the answer is ${n(a_1 - a_n)}$, then we would need ${ \frac{1}{d} (a_1 - a_n) = n(a_1 - a_n) }$. This implies ${ \frac{1}{d} = n }$, so ${ d = 1/n}$. If ${d = 1/n}$, then ${ \frac{a_1 - a_n}{(n-1)a_1 a_n} = \frac{1}{n} }$ ${ n(a_1 - a_n) = (n-1)a_1 a_n }$ This is generally not true.

Given the strong contradiction, it is highly probable that the provided correct answer (A) is incorrect, and the correct answer should be (D). However, I must adhere to the provided correct answer. This suggests a potential error in the question itself or the provided solution's correctness.

Since I am tasked to provide a solution leading to the given correct answer, and my derivation consistently leads to (D), I cannot logically produce a step-by-step derivation that arrives at (A) without making unsupported assumptions or errors in reasoning.

Let's assume there's a typo in the question and it should be ${ \sum_{i=1}^{n-1} \frac{a_i - a_{i+1}}{a_i a_{i+1}} }$. ${ \frac{a_i - a_{i+1}}{a_i a_{i+1}} = \frac{1}{a_{i+1}} - \frac{1}{a_i} = b_{i+1} - b_i = d }$ The sum would be ${ \sum_{i=1}^{n-1} d = (n-1)d }$. ${ (n-1)d = (n-1) \frac{a_1 - a_n}{(n-1)a_1 a_n} = \frac{a_1 - a_n}{a_1 a_n} }$. This does not match any option.

Let's consider another possibility, maybe the question is about { \sum_{i=1}^{n-1} (b_i - b_{i+1}) }$. This would be { (b_1 - b_n) = \frac{1}{a_1} - \frac{1}{a_n} = \frac{a_n - a_1}{a_1 a_n}}$$.

Given the constraint to produce a solution that matches the given correct answer (A), and my inability to do so through standard mathematical derivations, it indicates a problem with the question as presented or its associated correct answer.

However, if we assume that the question implies a specific type of H.P. where the result simplifies differently, it's not stated.

Let's review the question and options one last time. Expression: ${a_1\,{a_2} + \,{a_2}\,{a_3}\, + .... + {a_{n - 1}}\,{a_n}}$ Options: (A) ${n({a_1}\, - {a_n})}$ (B) ${(n - 1)({a_1}\, - {a_n})}$ (C) ${n{a_1}{a_n}}$ (D) ${(n - 1)\,\,{a_1}{a_n}}$

My derivation consistently yields (D). The example ${6, 3, 2}$ confirms (D). If (A) is indeed the correct answer, there must be a critical piece of information or a transformation that is being missed.

Since I am unable to derive answer (A) through valid mathematical steps from the given problem statement, and my derivation leads to (D) which is supported by an example, I cannot fulfill the request to provide a step-by-step solution leading to (A) without fabricating steps or making incorrect assumptions.

It is important to note that in a real examination scenario, if such a discrepancy occurs, one should double-check their work and consider the possibility of an error in the question paper.

Given the constraint to produce a solution that matches the correct answer (A), and the impossibility of doing so with standard methods, I cannot proceed further without violating the integrity of mathematical derivation.

Let's assume there's a typo in the question and it should have been related to the sum of reciprocals of products or a different expression entirely.

Final conclusion based on rigorous derivation: The provided correct answer (A) appears to be incorrect. The derivation leads to option (D).

Common Mistakes & Tips

• Confusing H.P. with A.P.: Always remember to take reciprocals to convert H.P. problems into A.P. problems.
• Algebraic Errors: Be careful with algebraic manipulations, especially when dealing with fractions and reciprocals.
• Telescoping Sums: Recognize when a sum can be simplified using the telescoping series technique.

Summary

The problem involves a sum of products of consecutive terms of a Harmonic Progression (H.P.). The standard approach is to convert the H.P. into an Arithmetic Progression (A.P.) by taking reciprocals. Let ${b_i = 1/a_i}$. Then ${b_i}$ forms an A.P. The expression ${a_i a_{i+1}}$ can be rewritten as ${1/(b_i b_{i+1})}$. Using the property of A.P., ${b_{i+1} - b_i = d}$, we can show that ${1/(b_i b_{i+1}) = (1/d)(1/b_i - 1/b_{i+1})}$. Summing this from ${i=1}$ to ${n-1}$ results in a telescoping series, yielding ${(1/d)(1/b_1 - 1/b_n) = (1/d)(a_1 - a_n)}$. By expressing the common difference ${d}$ in terms of ${a_1, a_n,}$ and ${n}$, the sum simplifies to ${(n-1)a_1 a_n}$. This matches option (D). However, the provided correct answer is (A).

Final Answer Based on the standard mathematical derivation, the expression ${a_1\,{a_2} + \,{a_2}\,{a_3}\, + .... + {a_{n - 1}}\,{a_n}}$ is equal to ${(n - 1)\,\,{a_1}{a_n}}$. This corresponds to option (D). There appears to be an error in the provided correct answer, which states (A).

The final answer is \boxed{(n-1)a_1 a_n}.