Q83 of 84 Page 5

If the mth term of an A. P. is 1/n and nth term is 1/m then show that its (mn)th term is 1. (CBSE 2017)

Given that am = 1/n and an = 1/m

We have to show that amn = 1.


We know that in an AP with first term a and common difference d, the nth term (or the general term) is given by an = a + (n – 1) d.


∴ am = a + (m – 1) d and an = a + (n – 1) d


⇒ 1/n = a + (m – 1) d … (1)


⇒ 1/m = a + (n – 1) d … (2)


Solving (1) and (2),


⇒ (a – a) + (md – d – nd + d) = 1/n – 1/m


⇒ (m-n) d = 1/n – 1/m


⇒ (m – n) d = (m – n) / mn


∴ d = 1/mn


Substituting d value in (1),


⇒ a + (m – 1) (1/mn) = 1/n


⇒ a + 1/n – 1/mn = 1/n


⇒ a = 1/mn + 1/n – 1/n


∴ a = 1/mn


Now, amn = a + (mn – 1) d


Substituting a and d values,


⇒ amn = (1/mn) + (mn – 1) (1/mn)


⇒ amn = (1/mn) + 1 – (1/mn)


∴ amn = 1


Ans. Hence from the given conditions we showed that amn = 1.

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