Sum of pqth Term in Arithmetic Progression

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Discussion Overview

The discussion revolves around finding the sum of the pqth term in an arithmetic progression (AP) given that the pth term is 1/q and the qth term is 1/p. Participants explore the relationships between the terms and the implications for the sum.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant states the relationship for the pth and qth terms of the AP as equations involving the first term and common difference.
  • Another participant derives that both the first term and the common difference equal 1/pq, but expresses confusion regarding the sum of the pqth term.
  • A later reply confirms the earlier finding about the first term and common difference and suggests applying the sum formula for the first n terms of an AP.
  • Finally, a participant calculates the sum of the pqth term using the derived values and presents the result as pq.

Areas of Agreement / Disagreement

Participants generally agree on the relationships derived from the definitions of the terms in the arithmetic progression, but there is confusion expressed regarding the sum of the pqth term, indicating some uncertainty in the calculations.

Contextual Notes

There are unresolved aspects regarding the assumptions made about the arithmetic progression and the implications of the derived values on the sum calculation.

Doffy
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The pth term of an airthmetic progression is 1/q and qth term is 1/p. What is the sum of pqth term?
 
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I would begin by stating:

$$a_p=a_1+(p-1)d=\frac{1}{q}$$

$$a_q=a_1+(q-1)d=\frac{1}{p}$$

Now, solve both for $a_1$...then equate the results and solve that for $d$...what do you find?
 
According to this, we find that a1=d=1/pq.
However, I am still confused about the sum of pqth term.
 
Doffy said:
According to this, we find that a1=d=1/pq.
However, I am still confused about the sum of pqth term.

Yes, that's correct. So now apply the formula:

$$S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)$$

where:

$$n=pq,\,a_1=d=\frac{1}{pq}$$
 
What we then get is:

$$S_{pq}=\frac{pq}{2}\left(\frac{2}{pq}+\frac{pq-1}{pq}\right)=1+pq-1=pq$$
 

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