Question

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

Answer 1

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

Complete step by step solution:
We are given that ${S_m} = {S_n}$
Let the first term of the A.P. is $a$ and let the common difference of the A.P. is $d$, then
${S_m} = \dfrac{m}{2}\left( {2a + \left( {m - 1} \right)d} \right)$ and
${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
From the given condition,
$\dfrac{m}{2}\left( {2a + \left( {m - 1} \right)d} \right) = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)0$
On simplifying it whether, we will get,
$
  m\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 = \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, ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
$
  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 ${S_{m + n}}$ which is also equal to ${S_{m + n}} = \dfrac{{m + n}}{2}\left( {2a + \left( {m + n - 1} \right)d} \right)$
We will now substitute the value of $2a$ in the above equation.
$
  {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 ${S_{m + n}}$ is 0.

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