Question #6889

In an Arithmetic Progression(AP) if Sm=Sn, then prove that Sm+n=0

Expert's answer

The sum of the first n terms of an arithmetic progression can be expressed in a such way:

S(n) = n*(a(1)+a(n))/2,

where a(1) is the initial term of an arithmetic progression and a(n) = a(1) + (n-1)d is the n-th term of the progression.

Let's consider equality S(n) = S(m):

S(n) = n*(a(1)+a(n)) = m*(a(1)+a(n)) = S(m)& ==>

a(1)*(n-m) = a(n)*(m-n)& ==>

a(1) = -a(n)& ==>

a(1) = -(a(1)+(n-1)*d)& ==>

n=1: 2*a(1)=0& ==> a(1)=0;

n=2: 2*a(2)=d& ==> d=0;

a(1)=d=0& ==> S(m)=S(n)=S(m)=S(m+n)=0.

S(n) = n*(a(1)+a(n))/2,

where a(1) is the initial term of an arithmetic progression and a(n) = a(1) + (n-1)d is the n-th term of the progression.

Let's consider equality S(n) = S(m):

S(n) = n*(a(1)+a(n)) = m*(a(1)+a(n)) = S(m)& ==>

a(1)*(n-m) = a(n)*(m-n)& ==>

a(1) = -a(n)& ==>

a(1) = -(a(1)+(n-1)*d)& ==>

n=1: 2*a(1)=0& ==> a(1)=0;

n=2: 2*a(2)=d& ==> d=0;

a(1)=d=0& ==> S(m)=S(n)=S(m)=S(m+n)=0.

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