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Let mth term be am and nth term be an
According to the question :
m(am ) = n(an )
Apply Formula : an = a + (n-1) *d [ a - first term , d -Common Difference , n - No of terms ]
m( a + (m-1) *d ) = n( a + (n-1) *d )
{ma + (m2 - m) d }= [na + (n2 - n) d]
{ma + (m2 d- md) }= [na + (n2 d- nd) ]
[ma-na] + { m2 d- n2 d } + (nd-md) = 0
a*[m-n] + d* { m2 - n2 } + d* (n-m) = 0
Apply : a2 - b2 = (a+b) (a-b)
a*[m-n] + d* { (m+n) (m-n) } - d* (m - n) = 0
Divide the above equation with " (m-n)" ,We get :
a + d* { (m+n) } - d = 0
a + { (m+n) - 1 } d = 0
COMPARING THE EQUATION WITH : an = a + (n-1) *d
Therefore : a + { (m+n) - 1 } d = am+n
So : am+n = 0
So :the (m+n)th term of the A.P is zero
Regards
Arun (askIITians forum expert)
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