Question

if the term of an ap is 1 /n and nth term is 1/ m then show that its (MN )th term is 1

Answer 1

Am=1/n
a+(m-1)d=1/n
a+md-d=1/n. 1
An=1/m
a+(n-1)d=1/m
a+nd-d=1/m. 2
1-2
(m-n)d=1/n-1/m
(m-n)d=m-n/mn
d=1/mn
Putting d=1/mn in eq.1
a+m(1/mn)-1/mn=1/n
a=1/mn
Amn=a+(mn-1)d
Amn=1/mn+mn/mn-1/mn
Amn=mn/mn
Amn=1
Hence proved

Answer 2

Hey..

Here's your answer..

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$am = 1 \div n \\ an = 1 \div m \\ \\ so....a + (m - 1)d = 1 \div n \\ a(n - 1)d = 1 \div m \\ \\ subtracting \: equtions \: we \: get \: ... \\ d(m - 1 - n + 1) = m - n \div mn \\ d(m - n) = m - n \div mn \\ \\ \\ thus \: d = 1 \div mn \\ \\ \\ now \: putting \: value \: of \: d \: in \: any \: of \: these \: equation \\ \\ a \: will \: be \: 1 \div mn$
$now \: a(mn) = a + (mn - 1)d \\ \\ 1 \div mn + mn \div mn - 1 \div mn \\ = 1$
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HOPE IT HELPS..

@Rêyaañ11