Question

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.

Answer 1

We know :- a_n = a +(n-1)d
a _(m+n) = a + (m+n-1)d (just put m+n in place of n ) --------------(1)
Let the first term and common difference of the A.P. be ‘a’ and ‘d’ respectively.
Then, m ^th term = a + (m – 1) d and n ^th term = a + (n-1)d
By the given condition,
m x a_m = n x a_n
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 ---------------(ii)
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