Question

In arithmetic progression , Tm=n and Tn=m than prove that mn th term is 1.how to solve this ?

Answer 1

I think it is m+n th term
Let a = first term of AP and d the common difference
We have Tm= n => a+(m-1)d = n (1)
Also Tn=m => a+(n-1)d =m (2)
Subtract (2) from (1) giving a+md-d-a-nd+d=n-m=>md-nd=n-m=>(m-n)d=-(m-n) => d= -1
Substitute d = -1 into (1) giving
a-(m-1) = n=> a-m+1= n => a = m+n-1 (3)
Tm+n = a+(m+n-1)d = a-(m+n-1) since d = -1
= a-m-n+1 Now substitute a = m+n-1 from (3)
Tm+n = m+n-1-m-n+1 = 0

(m+n) th term is zero