it is given that mth term is equal to nth term.therefore,
m am = n an
using an formula a +(n-1)d,
m [a(m-1)d] = n [a + (n-1)d]
Multiply the equation with m and n,
ma + (m2-m)d = na + (n2 - n)d
bring both equations to 1 side,
ma - na + (m2-m)d - (n2-n)d =0
take common factor "a and d" out,
(m-n)a + (m2 - m -n2 +n )d = 0
bring square terms together,
(m-n)a + (m2 - n2- m - n)d =0
Now, it is in the for a2-b2 write it as (a+b) (a-b)
(m-n)a [(m+n)(m-n) - (m-n)] d =0
take (m-n) out as common,
(m-n) [a+(m+n-1)d] =0
(m-n) can be taken to the other side and it becomes 0,
a+ (m+n-1)d =0
the final equation is obtained from the an formula a+(n-1)d
am + an =0
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