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26 Apr 2012 11:11:08 IST
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Here mth term is Am = n
and nth term is An = m
Let first term = a
and common difference = d
Thus Am = n = a + (m -1) d ...(1)
An = m = a + (n-1) 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 +(m-1)d=n......................(1)
a +(n-1)d=m.........................(2)
substracting eq 1 from 2 we get:
d(m-n)=n-m
d=-(m-n)/m-n
d=-1.......................................(3)
putting the value of d in eq 1 we get
a +(m-1)-1=n
a=n+m-1...........................(4)
Sm+n=m+n/2 *2(m+n-1)+(m+n-1)-1
Sm+n=(m+n)(m+n-1)/2
and am+n=a + (m+n-1)-1
(puting the value of a from eq4 we get)
an=(m+n-1) - (m+n-1)
an=0