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Here mth term is A_{m} = n
and nth term is A_{n} = m
Let first term = a
and common difference = d
Thus Am = n = a + (m 1) d ...(1)
An = m = a + (n1) d ...(2)
Solving equations (1) and (2) simultaneously we obtain
d = 1
and a = n + m  1
Thus sum of (m + n) terms is
S_{(m + n)} = [(n + m) / 2] * [2a + ( n + m  1)d]
or S_{(m + n)} = [(n + m) / 2] * [2(n + m  1)  ( n + m  1)]
or S_{(m + n)} = (n + m) * (n + m  1) / 2
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a +(m1)d=n......................(1)
a +(n1)d=m.........................(2)
substracting eq 1 from 2 we get:
d(mn)=nm
d=(mn)/mn
d=1.......................................(3)
putting the value of d in eq 1 we get
a +(m1)1=n
a=n+m1...........................(4)
S_{m+n}=m+n/2 *2(m+n1)+(m+n1)1
S_{m+n}=(m+n)(m+n1)/2
and a_{m+n}=a + (m+n1)1
(puting the value of a from eq4 we get)
a_{n}=(m+n1)  (m+n1)
a_{n}=0