In an A.P. if $p$th term is $\dfrac{1}{q}$ and $q$th term is $\dfrac{1}{p}$, prove that the sum of first $pq$ terms is $\dfrac{1}{2}(pq+1)$.

> [!TLDR] Using the two given term conditions we find $a=d=\dfrac1{pq}$, and substituting into the sum formula proves $S_{pq}=\dfrac12(pq+1)$. > [!FORMULA] > $n$th term: $a_n=a+(n-1)d$. Sum of $n$ terms: $S_n=\dfrac n2\big[2a+(n-1)d\big]$. 1. Given $a_p=a+(p-1)d=\dfrac1q\ \ldots(i)$ and $a_q=a+(q-1)d=\dfrac1p\ \ldots(ii)$. 2. Subtract (ii) from (i): $(p-q)d=\dfrac1q-\dfrac1p=\dfrac{p-q}{pq}$. 3. Since $p\ne q$, divide both sides by $(p-q)$: $d=\dfrac1{pq}$. 4. Substitute $d$ back into (i): $a=\dfrac1q-(p-1)\cdot\dfrac1{pq}=\dfrac{p-(p-1)}{pq}=\dfrac1{pq}$. 5. So $a=d=\dfrac1{pq}$. Now compute $S_{pq}=\dfrac{pq}{2}\big[2a+(pq-1)d\big]=\dfrac{pq}{2}\left[\dfrac{2}{pq}+\dfrac{pq-1}{pq}\right]=\dfrac{pq}{2}\cdot\dfrac{pq+1}{pq}=\dfrac12(pq+1)$. …

In an A.P. if pth term is (1)/(q) and qth term is (1)/(p), prove that the sum of first pq terms is (1)/(2)(pq+

Exercise 5.1 · Q10

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Concept understanding — Introduction

Introduction — What It Means and Why It Matters

You've been solving problems your whole life. When you see a new type of question, you first figure out what it's asking, then decide which tools to use, then carry out the steps. That first step — figuring out what's going on — is the heart of an introduction in mathematics and science.

An introduction is not a summary. It's a setup. It tells you the players, the rules, and the goal before the action starts.

The Intuition

Imagine you walk into a room and see a chessboard with pieces already arranged. Without knowing whose turn it is, what the last move was, or whether you're playing standard chess or a variant, you can't make sense of anything. An introduction is like being told: "White to move, standard rules, checkmate in two."

In a proof, an introduction declares the assumptions you're allowed to use. In a problem, it states the given data and the unknown you're after. In a definition, it names the new object and lists its essential properties.

Note

The word "introduction" here is about logical introduction — bringing a new idea or assumption into a discussion — not the first paragraph of an essay.

The Precise Statement

In formal logic and mathematics, an introduction rule tells you how to legitimately bring a new statement into a derivation. The most common example is the conditional introduction (also called "direct proof" or "→-introduction"):

If from assumption $P$ you can derive $Q$ , then you may conclude $P \implies Q$ .

That is, to prove "if $P$ then $Q$ ", you temporarily assume $P$ is true, work through the reasoning, and arrive at $Q$ . Then you discharge the assumption and write $P \implies Q$ as a proven statement.

Symbolically:

\frac{ \begin{array}{c} P \\ \vdots \\ Q \end{array} }{ P \implies Q }

The line means "from the top, infer the bottom." The dots represent a valid chain of reasoning.

Why This Matters for Exams

When you see a question that asks you to "prove that if $x > 2$ then $x^2 > 4$ ", you are being asked to perform a conditional introduction. You start by introducing the assumption $x > 2$ , work from there, and end with $x^2 > 4$ . Then you write the implication. …

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Using the two given term conditions we find $a=d=\dfrac1{pq}$ , and substituting into the sum formula proves $S_{pq}=\dfrac12(pq+1)$ .

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Given $a_p=a+(p-1)d=\dfrac1q\ \ldots(i)$ and $a_q=a+(q-1)d=\dfrac1p\ \ldots(ii)$ .
Subtract (ii) from (i): $(p-q)d=\dfrac1q-\dfrac1p=\dfrac{p-q}{pq}$ .
Since $p\ne q$ , divide both sides by $(p-q)$ : $d=\dfrac1{pq}$ .
Substitute $d$ back into (i): $a=\dfrac1q-(p-1)\cdot\dfrac1{pq}=\dfrac{p-(p-1)}{pq}=\dfrac1{pq}$ .
So $a=d=\dfrac1{pq}$ . Now compute $S_{pq}=\dfrac{pq}{2}\big[2a+(pq-1)d\big]=\dfrac{pq}{2}\left[\dfrac{2}{pq}+\dfrac{pq-1}{pq}\right]=\dfrac{pq}{2}\cdot\dfrac{pq+1}{pq}=\dfrac12(pq+1)$ . …

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Q.In an A.P. if pppth term is 1q\dfrac{1}{q}q1​ and qqqth term is 1p\dfrac{1}{p}p1​, prove that the sum of first pqpqpq terms is 12(pq+1)\dfrac{1}{2}(pq+1)21​(pq+1).

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Q.In an A.P. if $p$ th term is $\dfrac{1}{q}$ and $q$ th term is $\dfrac{1}{p}$ , prove that the sum of first $pq$ terms is $\dfrac{1}{2}(pq+1)$ .

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