QUESTION

# If $m$ times the ${{m}^{th}}$ term of an A.P. is equal to $n$ times its ${{n}^{th}}$ term, then show that the ${{\left( m+n \right)}^{th}}$ term of the A.P. is zero.

Hint: First of all, write the ${{m}^{th}}$ and the ${{n}^{th}}$ term of A.P. by using the formula, ${{a}_{n}}=a+\left( n-1 \right)d$. Now equate $n$ times of ${{n}^{th}}$ term to $m$ times the ${{m}^{th}}$ term. From here find the value of $a$ and substitute it in ${{\left( m+n \right)}^{th}}$ term to get the desired result.

We are given that if $m$ times the ${{m}^{th}}$ term of an A.P. is equal to $n$ times its ${{n}^{th}}$ term, then we have to show that the ${{\left( m+n \right)}^{th}}$ term of the A.P. is zero. Before proceeding with the question, let us see what an A.P. is. An A.P. or Arithmetic Progression is a sequence of numbers such that the difference of any two successive terms is constant. For example, the sequence of 2, 4, 6, 8 is an A.P. with the common difference of 2. We generally write the terms of A.P. as: $a,a+d,a+2d,a+3d\ldots$, where $a$ is the first term and $d$ is the common difference of A.P. Also, the ${{n}^{th}}$ term of A.P. is given by, ${{a}_{n}}=a+\left( n-1 \right)d$.
Now let us consider our question. We know that the ${{n}^{th}}$ term of A.P. is given by, ${{a}_{n}}=a+\left( n-1 \right)d\ldots \ldots \ldots \left( i \right)$.
So, by substituting $n=m$ in this equation, we get, ${{m}^{th}}$ term of A.P. as, $\left( {{a}_{m}} \right)=a+\left( m-1 \right)d$. By multiplying on both sides of the equation, we get,
$m$ times the ${{m}^{th}}$ term of an A.P. $\Rightarrow m{{a}_{m}}=m\left[ a+\left( m-1 \right)d \right]\ldots \ldots \ldots \left( ii \right)$
Also, by multiplying $n$ to both sides of equation (i), we get,
$n$ times the ${{n}^{th}}$ term of an A.P. $\Rightarrow n{{a}_{n}}=n\left[ a+\left( n-1 \right)d \right]\ldots \ldots \ldots \left( iii \right)$
We are given that $m$ times the ${{m}^{th}}$ term of an A.P. is equal to $n$ times of the ${{n}^{th}}$ term. So, by equating equation (ii) and equation (iii), we get, $m\left[ a+\left( m-1 \right)d \right]=n\left[ a+\left( n-1 \right)d \right]$. Simplifying this equation, we get,
\begin{align} & ma+m\left( m-1 \right)d=na+n\left( n-1 \right)d \\ & \Rightarrow \left( ma-na \right)=n\left( n-1 \right)d-m\left( m-1 \right)d \\ & \Rightarrow a\left( m-n \right)=d\left[ n\left( n-1 \right)-m\left( m-1 \right) \right] \\ & \Rightarrow a\left( m-n \right)=d\left[ {{n}^{2}}-n-{{m}^{2}}+m \right] \\ \end{align}
We can also write the above equation as,
$a\left( m-n \right)=d\left[ \left( m-n \right)-\left( {{m}^{2}}-{{n}^{2}} \right) \right]$
We know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. By using this in the above equation, we get, $a\left( m-n \right)=d\left[ \left( m-n \right)-\left( m-n \right)\left( m+n \right) \right]$
By taking out $\left( m-n \right)$, the common term from RHS of the above equation, we get,
$a\left( m-n \right)=\left( m-n \right)d\left[ 1-\left( m+n \right) \right]$
By cancelling $\left( m-n \right)$ from both the sides, we get, $a=d\left[ 1-m-n \right]\ldots \ldots \ldots \left( iv \right)$. By substituting $n=m+n$ in equation (i), we get, ${{\left( m+n \right)}^{th}}$ term of the A.P. as ${{a}_{m+n}}=a+\left( m+n-1 \right)d$. By substituting the value of $a$ from equation (iv) in this equation, we get, ${{a}_{m+n}}=d\left[ 1-m-n \right]+\left( m+n-1 \right)d$. By simplifying, we get,
\begin{align} & {{a}_{m+n}}=d-dm-dn+dm+dn-d \\ & \Rightarrow {{a}_{m+n}}=0 \\ \end{align}
Hence, we have proved that ${{\left( m+n \right)}^{th}}$ term of the A.P. is zero.

Note: In this question, students must note that the value of $a$ and $d$ will remain constant as we are talking about the same A.P. throughout the question, while the value of the ${{n}^{th}}$ term will change according to the information provided in the question. Also, students can substitute $d=\dfrac{a}{1-\left( m+n \right)}$ in the equation for ${{m}^{th}}$ term to prove the desired answer, but it is advisable to take the numbers or the terms in whole number values instead of fractions to avoid any mistakes.