if the mth term of an ap is 1/n and the nth term is 1/m, then prove that amn=1
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Tm=a+(m-1) d=1/n-------(1)
Tn=a+(n-1) d=1/m-------(2)
where a is first term and d is common difference
subtracting (1) to (2)
(m-n) d=1/n - 1/m
(m-n) d=(m-n)/mn
d=1/mn
and a=1/n - 1/mn (m-1)=1/mn
now
Tmn=a+(mn-1) d
=1/mn+(mn-1) 1/mn=1
hence answer is 1
Tn=a+(n-1) d=1/m-------(2)
where a is first term and d is common difference
subtracting (1) to (2)
(m-n) d=1/n - 1/m
(m-n) d=(m-n)/mn
d=1/mn
and a=1/n - 1/mn (m-1)=1/mn
now
Tmn=a+(mn-1) d
=1/mn+(mn-1) 1/mn=1
hence answer is 1
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