if m times the mth term of an AP is equal to n times the nth term . show that (m+n)th term of the AP IS ZERO
We know :- an = a +(n-1)d
a (m+n) = a + (m+n-1)d (just put m+n in place of n ) (mark it equation number 1)
Let the first term and common difference of the A.P. be ‘a’ and ‘d’ respectively.
Then, mth term = a + (m – 1) d and nth term = a + (n – 1) d
By the given condition,
m x am = n x an
m [a + (m – 1) d] = n [a + (n – 1) d]
⇒ ma + m (m – 1) d = na + n (n – 1) d
=> ma + (m2 -m)d - na - (n2 -n)d = 0 ( taking the Left Hand Side to the other side )
=> ma -na + (m2 - m)d -( n2-n)d = 0 (re-ordering the terms)
=> a (m-n) + d (m2-n2-m+n) = 0 (taking 'a' and 'd' common)
=> a (m-n) + d {(m+n)(m-n)-(m-n)} = 0 (a2-b2 identity)
Now divide both sides by (m-n)
=> a (1) + d {(m+n)(1)-(1)} = 0
=>a + d (m+n-1) = 0 (mark it equation number 2)
From equation number 1 and 2 ,
a (m+n) = a + (m+n-1)d
And we have shown ,
a + d (m+n-1) = 0
So, a (m+n) = 0 (Answer!!!!!)
this one was really tough :O