If the mth term of an Ap is 1/n and nth term is 1/m then show that its (mn)th term is 1.
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given that, mth term=1/n and nth term=1/m.
then ,let a and d be the first term and the common difference of the A.P.
so a+(m-1)d=1/n...........(1) and a+(n-1)d=1/m...........(2).
subtracting equation (1) by (2) we get,
md-d-nd+d=1/n-1/m
=>d(m-n)=m-n/mn
=>d=1/mn.
again if we put this value in equation (1) or (2) we get, a=1/mn.
then, let A be the mnth term of the AP
a+(mn-1)d=1/mn+1+(-1/mn)=1
hence proved
then ,let a and d be the first term and the common difference of the A.P.
so a+(m-1)d=1/n...........(1) and a+(n-1)d=1/m...........(2).
subtracting equation (1) by (2) we get,
md-d-nd+d=1/n-1/m
=>d(m-n)=m-n/mn
=>d=1/mn.
again if we put this value in equation (1) or (2) we get, a=1/mn.
then, let A be the mnth term of the AP
a+(mn-1)d=1/mn+1+(-1/mn)=1
hence proved
Answered by
2
Answer:
given that, mth term=1/n and nth term=1/m.
then ,let a and d be the first term and the common difference of the A.P.
so a+(m-1)d=1/n...........(1) and a+(n-1)d=1/m...........(2).
subtracting equation (1) by (2) we get,
md-d-nd+d=1/n-1/m
=>d(m-n)=m-n/mn
=>d=1/mn.
again if we put this value in equation (1) or (2) we get, a=1/mn.
then, let A be the mnth term of the AP
a+(mn-1)d=1/mn+1+(-1/mn)=1
hence proved.
Step-by-step explanation:
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