Sum of pqth Term in Arithmetic Progression

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SUMMARY

The discussion focuses on calculating the sum of the pqth term in an arithmetic progression (AP) where the pth term is 1/q and the qth term is 1/p. The initial terms are defined as \(a_p = a_1 + (p-1)d = \frac{1}{q}\) and \(a_q = a_1 + (q-1)d = \frac{1}{p}\). By solving these equations, it is established that both \(a_1\) and \(d\) equal \(\frac{1}{pq}\). The sum of the pqth term is derived using the formula \(S_n = \frac{n}{2}(2a_1 + (n-1)d)\), resulting in \(S_{pq} = pq\).

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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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