# If the mth term of an A.P. be 1/n and nth term be 1/m, then show that its (mn)th term is 1. - Mathematics

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

#### Solution 1

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

Then,

1/n= mth term ⇒ 1/n = a + (m – 1) d ….(i)

1/m = nth term ⇒ 1/m = a + (n – 1) d ….(ii)

On subtracting equation (ii) from equation (i), we get

\frac{1}{n}-\frac{1}{m}=( mn)d

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

\frac{1}{n}=a+\frac{(m-1)}{mn}\Rightarrow a=\frac{1}{mn}

∴ (mn)th term = a + (mn – 1) d

=\frac{1}{mn}+(mn-1)\frac{1}{mn}=1

#### Solution 2

Let a and d be the first term and common difference respectively of the AP, respectively.
Then,

mth term =1/n

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

a = 1/n - (m -1)d .....(1)

"nth term"= 1/m

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

=> 1/n -(m - 1)d + (n - 1)d =   1/m   [From 1]

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

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

=> d = 1/"mn"

Putting d = 1/mn in (1) we have

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

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

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

:. t_"mn" = a + (mn - 1)d = 1/(mn) + (mn -1) 1/(mn) = 1

Concept: Arithmetic Progression
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