Find the $(mn)^{th}$ term

Question

Here is a question from an IB past paper -

If the $m^{th}$ term of an Arithmetic Sequence is $\frac{1}{n}$ and $n^{th}$ term be $\frac{1}{m}$ then find its $(mn)^{th}$ term.

My working -

$$a+(n-1)d=\frac{1}{m}$$

$$a+(m-1)d=\frac{1}{n}$$

$$\Rightarrow\;(n-1)d-(m-1)d=\frac{1}{m}-\frac{1}{n}$$

$$\Rightarrow\;(n-m)d=\frac{n-m}{mn}$$

$$\Rightarrow\;(mn)d=\frac{n-m}{n-m}$$

$$\therefore\;(mn)d=1$$

Is this the right answer? If not, what is the $(mn)^{th}$ term?

Answer 1

Yes, it is right. You have $d=\frac{1}{mn}$. Similarly, isolate $a$ in the first two equations by multiplying $(m-1)$ in the first equation and $(n-1)$ in the second equation. You can work out that $a=\frac{1}{mn}$. Then, the $(mn)^{th}$ term would be: $$a+(mn-1)(d)=\frac{1}{mn}+\frac{mn-1}{mn}=1$$

Answer 2

Useful formula to remember: $$d=\frac{a_m-a_n}{m-n}=\frac{\frac1n-\frac1m}{m-n}=\frac1{mn}\\ d=\frac{a_{mn}-a_n}{mn-n}=\frac{a_{mn}-\frac1m}{mn-n} \Rightarrow \\ a_{mn}=\frac1m+\frac1{mn}(mn-n)=1.$$