If Mth term of an A.P. …
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Gaurav Seth 6 years, 11 months ago
Let a and d be the first term and common difference respectively of the given AP,then
mth term = a+(m-1)d ==> a+(m-1)d = 1/n -----(1)
and nth term=a+(n-1)d ==> a+(n-1)d = 1/m -----(2)
subtracting (2) from (1), we have
Let the first term of the A.P be 'a ' and its common difference be 'd '
Now, mth term of the A.P is 1/n
which means, a + (m - 1)d = 1/n --- (i)
Again,
its nth term is 1/m
which means, a + (n - 1)d = 1/m --- (ii)
on substracting eqn. (ii) from (i) we get,
a + (m - 1)d - a - (n - 1)d = 1/n - 1/m
=> d (m - 1 - n + 1) = (m - n)/mn
=> d (m - n) = (m - n) / mn
=> d = 1/mn
so, a = 1/m - (n - 1)(1/mn) = n - n + 1 / mn =1/mn
ie., a = d = 1/mn
Now, Smn = mn/2 [2a + (mn - 1)d]
= mn/2 [2/mn + (mn - 1)1/mn]
= mn/2 [2 + mn - 1]/mn
= (mn + 1)/2
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