Question

If the m th term of an AP is 1/n and n th term is 1/m then show that its (mn) th term is 1.

Answer 1

The answer is given below :

Let us consider that the first term of the AP is a and the common ratio is d.

Given,

m-th term = 1/n

=> a + (m - 1)d = 1/n .....(i)

and

n-th term = 1/m

=> a + (n - 1)d = 1/m .....(ii)

We have

a + (m - 1)d = 1/n .....(i)
a + (n - 1)d = 1/m .....(ii)

On subtraction, we get

(m - 1 - n + 1)d = (1/n - 1/m)

=> (m - n)d = (m - n)/mn

=> d = 1/mn

So, common ratio = 1/mn

Putting d = 1/mn in (i), we get 

a + (m - 1)(1/mn) = 1/n

=> a + 1/n - 1/mn = 1/n 

=> a = 1/mn [cancelling 1/n from both sides]

So, first term = 1/mn

Therefore, the mn-th term is

= a + (mn - 1)d

= 1/mn + (mn - 1)(1/mn)

= 1/mn + mn/mn - 1/mn

= 1 [Proved]

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