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Question: If \({S_m} = {S_n}\) for some A.P, then prove that \({S_{m + n}} = 0\)....

If Sm=Sn{S_m} = {S_n} for some A.P, then prove that Sm+n=0{S_{m + n}} = 0.

Explanation

Solution

We will write the sum of first mm and nn terms of A.P. and then put them equal. Simplify the expression and find the value of 2a2a. Then, write the expression of Sm+n{S_{m + n}} and substitute the value of 2a2a and solve the equation further.

Complete step by step solution:
We are given that Sm=Sn{S_m} = {S_n}
Let the first term of the A.P. is aa and let the common difference of the A.P. is dd, then
Sm=m2(2a+(m−1)d){S_m} = \dfrac{m}{2}\left( {2a + \left( {m - 1} \right)d} \right) and
Sn=n2(2a+(n−1)d){S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)
From the given condition,
m2(2a+(m−1)d)=n2(2a+(n−1)d)0\dfrac{m}{2}\left( {2a + \left( {m - 1} \right)d} \right) = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)0
On simplifying it whether, we will get,
m(2a+(m−1)d)=n(2a+(n−1)d) ⇒2am+(m−1)md=2an(n−1)md m\left( {2a + \left( {m - 1} \right)d} \right) = n\left( {2a + \left( {n - 1} \right)d} \right) \\\ \Rightarrow 2am + \left( {m - 1} \right)md = 2an\left( {n - 1} \right)md
Now, we will put like terms together,
2am−2an=(n−1)nd−(m−1)md ⇒2a(m−n)=d(n2−n−m2+m) ⇒2a(m−n)=d(n2−m2−n+m) 2am - 2an = \left( {n - 1} \right)nd - \left( {m - 1} \right)md \\\ \Rightarrow 2a\left( {m - n} \right) = d\left( {{n^2} - n - {m^2} + m} \right) \\\ \Rightarrow 2a\left( {m - n} \right) = d\left( {{n^2} - {m^2} - n + m} \right)
Apply the formula, a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)
2a(m−n)=d(−(m+n)(m−n)−(m+n)) ⇒2a(m−n)=d(m−n)(1−m−n) ⇒2a=d(1−m−n) 2a\left( {m - n} \right) = d\left( { - \left( {m + n} \right)\left( {m - n} \right) - \left( {m + n} \right)} \right) \\\ \Rightarrow 2a\left( {m - n} \right) = d\left( {m - n} \right)\left( {1 - m - n} \right) \\\ \Rightarrow 2a = d\left( {1 - m - n} \right)
We have to find the value of Sm+n{S_{m + n}} which is also equal to Sm+n=m+n2(2a+(m+n−1)d){S_{m + n}} = \dfrac{{m + n}}{2}\left( {2a + \left( {m + n - 1} \right)d} \right)
We will now substitute the value of 2a2a in the above equation.
Sm+n=m+n2(d(1−m−n)+(m+n−1)d) ⇒Sm+n=m+n2(d−dm−dn+dm+dn−d) ⇒Sm+n=m+n2(0) ⇒Sm+n=0 {S_{m + n}} = \dfrac{{m + n}}{2}\left( {d\left( {1 - m - n} \right) + \left( {m + n - 1} \right)d} \right) \\\ \Rightarrow {S_{m + n}} = \dfrac{{m + n}}{2}\left( {d - dm - dn + dm + dn - d} \right) \\\ \Rightarrow {S_{m + n}} = \dfrac{{m + n}}{2}\left( 0 \right) \\\ \Rightarrow {S_{m + n}} = 0

Hence, the sum of Sm+n{S_{m + n}} is 0.

Note:
In an A.P., the sequence is of type a,a+d,a+2d,a+3d,......a,a + d,a + 2d,a + 3d,....... The nth{n^{th}} term of the sequence is given as an=a+(n−1)d{a_n} = a + \left( {n - 1} \right)d. And the sum of nn terms of sequence is given by n2(2a+(n−1)d)\dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right) and n2(a+an)\dfrac{n}{2}\left( {a + {a_n}} \right)