Question

if mth term of ap is 1/n and nth term is a/ 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]

Thank you for your question.

Answer 2

AnswEr:

Let a and d be the term and common difference respectively of the given A.P. Then,

$\tt \frac{1}{n} = mth \: term \\ \\ \implies \tt \frac{1}{n} = a + (m - 1)d - - - (1)$

_____________

$\tt \frac{1}{m} = nth \: term \\ \\ \implies \tt \frac{1}{m} = a + (n - 1)d - - - (2)$

$\underline\text{On\:subtracting\:(2)\:from\:(1)\:,we\:get}$

$\tt \frac{1}{n} - \frac{1}{m} = (m - n)d \\ \\ \implies \tt \frac{m - n}{mn} = (m - n)d \\ \\ \implies \tt \: d = \frac{1}{mn}$

$\underline\text{Putting\:d=1/mn\:in\:(1)\:we\:get}$

$\tt \frac{1}{n} = a + \frac{(m - 1)}{mn} \\ \\ \tt \implies \frac{1}{n} = a + \frac{1}{n} - \frac{1}{mn} \\ \\ \tt \implies \: a = \frac{1}{mn} \\ \\ \sf \therefore \: (mn)th \: term = a + (mn - 1)d = \frac{1}{mn} + (mn - 1) \frac{1}{mn} = 1$