If m times the mth term of an A.P. is equal to n times the nth term, prove that (m+n)th term of A.P is zero

Arithmetic Progression is a sequence in which the difference between any two consecutive terms is always the same. In simple terms, it means that the next number in the series is calculated by adding a fixed number i.e. common difference to the previous number in the series. 

For example, f(n): 1,3,5,7,….. will be called an Arithmetic Progression because the difference between any two consecutive terms in the series (a common difference which is denoted as ‘d’) is the same i.e.,

The difference between the second and first term (3-1=2) is the same as the difference between the third and fourth term (5-3=2) is the same as the difference between the fourth term and the third term (7-5=2). Hence f(n) is an arithmetic progression.

Here,

the first term (a)=1, common difference (d)=2, no of terms (n)=4

T₄=1+(4-1)×2

=>7

The General Term for AP is: The formula for the Nth term of an AP is 

T_n = a + ( n- 1) × d,

where,

a is the first term of the series,

n is the no of terms and, 

d is the common difference between the two consecutive terms of the series.

How to find the General Term of An Arithmetic Progression

Let the first term be T₁, the second term be T_{2 ……………}then the nth term be T_n

Let the first term be a and the common difference between two consecutive terms be d.

T₁=a+0×d, 

T₂= a+1×d, 

T₃=a+(3-1)×d,then,

T_n=a+(n-1)×d

Therefore,

N^th term of an arithmetic progression = a+(n-1)×d 

Statement: If m times the m^th term is equal to n times the n^th term of an A.P. prove that (m + n)^th term of A.P is zero.

Proof:

Here,

It is given that m times the m^th term is equal to n times the n^th term we get,

Therefore, 

N^th terms of an AP is = T_n= a+(n-1) d ————>(1)

M^th terms of an AP is = T_m= a+(m-1) d ———->(2)

According the  question,

                     m×T_m=n×T_n

Equating equation 1 and 2 we get,

=> m[a + (m − 1)d] = n[a + (n − 1)d]

=> m[a + (m − 1)d] − n[a + (n − 1)d] = 0 //Taking all terms to LHS we get,

=> a(m − n) + d[(m + n)(m − n) − (m − n)] = 0

=> (m − n)[a + d((m + n) − 1)] = 0 //dividing both sides by (m-n) we get,

=> a + [(m + n) − 1]d = 0          ——————->(3)

The above equation represents the (m+n)^th terms of the series T_m+n.

Thus,

=> T_m+n = 0

Hence above statement is Proved.