Sum of pqth Term in Arithmetic Progression

[Doffy]

The p^th term of an airthmetic progression is 1/q and q^th term is 1/p. What is the sum of pq^th term?

 

[MarkFL]

I would begin by stating:

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

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

Now, solve both for $a_1$...then equate the results and solve that for $d$...what do you find?

 

[Doffy]

According to this, we find that a₁=d=1/pq.
However, I am still confused about the sum of pq^th term.

 

[MarkFL]

According to this, we find that a₁=d=1/pq.
However, I am still confused about the sum of pq^th term.

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

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

where:

\(\displaystyle n=pq,\,a_1=d=\frac{1}{pq}\)

 

[MarkFL]

What we then get is:

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