Question

In an A.P, if $m\left( {{a_m}} \right) = n\left( {{a_n}} \right)$. Find $\left( {{a_m} + {a_n}} \right)th$ term.

Answer 1

Hint: In order to find the value of $\left( {{a_m} + {a_n}} \right)th$, we need to solve the given equation $m\left( {{a_m}} \right) = n\left( {{a_n}} \right)$, which represents that m times of the ${a_m}$value is equal to n times of the ${a_n}$ value, substitute the formula solve and get the results. Expand the term $\left( {{a_m} + {a_n}} \right)th$, substitute the values needed, solve and get the results.
Formula used:
\[{T_n} = {a_1} + \left( {n - 1} \right)d\]

Complete step-by-step solution:
We are given with an equation $m\left( {{a_m}} \right) = n\left( {{a_n}} \right)$.
Finding each term of the brackets separately, using the formula from AP, which is:
For, any term we write:
${a_x} = {a_1} + \left( {x - 1} \right)d$ ……(1)
Where ${a_1}$ is the first term, x is the ${x^{th}}$ term we need to find and $d$ is the common difference in the AP.
Now, we need to find ${a_m}$ term, so substituting m in places of x, in the equation 1, and we get:
${a_m} = {a_1} + \left( {m - 1} \right)d$
Multiplying both the sides by m:
$m\left( {{a_m}} \right) = m\left( {{a_1} + \left( {m - 1} \right)d} \right)$
Multiplying m on the right side with each value, we get:
$m\left( {{a_m}} \right) = m{a_1} + m\left( {m - 1} \right)d$
$ \Rightarrow m\left( {{a_m}} \right) = m{a_1} + \left( {{m^2} - m} \right)d$ ……..(2)
Similarly, we need to find ${a_n}$ term, so substituting n in places of x, in the equation 1, and we get:
${a_n} = {a_1} + \left( {n - 1} \right)d$
Multiplying both the sides by n:
$n\left( {{a_n}} \right) = n\left( {{a_1} + \left( {n - 1} \right)d} \right)$
Multiplying n on the right side with each value, we get:
$n\left( {{a_n}} \right) = n{a_1} + n\left( {n - 1} \right)d$
$ \Rightarrow n\left( {{a_n}} \right) = n{a_1} + \left( {{n^2} - n} \right)d$ ……..(3)
Since, we were given that $m\left( {{a_m}} \right) = n\left( {{a_n}} \right)$:
Subtracting both sides by $n\left( {{a_n}} \right)$, and we get:
$m\left( {{a_m}} \right) - n\left( {{a_n}} \right) = n\left( {{a_n}} \right) - n\left( {{a_n}} \right)$
$ \Rightarrow m\left( {{a_m}} \right) - n\left( {{a_n}} \right) = 0$
Substituting the values from equation 2 and equation 3 in the above equation and we get:
$ \Rightarrow m\left( {{a_m}} \right) - n\left( {{a_n}} \right) = m{a_1} + \left( {{m^2} - m} \right)d - n{a_1} + \left( {{n^2} - n} \right)d$
$ \Rightarrow m{a_1} + \left( {{m^2} - m} \right)d - n{a_1} + \left( {{n^2} - n} \right)d = 0$
Taking ${a_1}$ and d common:
$ \Rightarrow \left( {m - n} \right){a_1} + \left( {{m^2} - m - {n^2} + n} \right)d = 0$
Subtracting both sides by $\left( {{m^2} - m - {n^2} + n} \right)d$:
$ \Rightarrow \left( {m - n} \right){a_1} + \left( {{m^2} - m - {n^2} + n} \right)d - \left( {{m^2} - m - {n^2} + n} \right)d = 0 - \left( {{m^2} - m - {n^2} + n} \right)d$
$ \Rightarrow \left( {m - n} \right){a_1} = - \left( {{m^2} - m - {n^2} + n} \right)d$
Factoring $\left( {{m^2} - m - {n^2} + n} \right)$, and we can write it as: $\left( {{m^2} - m - {n^2} + n} \right) = \left( {m - n} \right)\left( {m + n - 1} \right)$, so we get:
$ \Rightarrow \left( {m - n} \right){a_1} = - \left( {m - n} \right)\left( {m + n - 1} \right)d$
Cancelling common terms from both the sides:
$ \Rightarrow {a_1} = - \left( {m + n - 1} \right)d$ …….(4)
We got the value of the first term of the AP.
Now, we need to find the value of $\left( {{a_m} + {a_n}} \right)th$ term. So, we can write it using equation 2. Replacing x with $\left( {m + n} \right)$, and we get:
$\left( {{a_m} + {a_n}} \right) = {a_1} + \left( {m + n - 1} \right)d$
Substituting the value of ${a_1}$, from equation 4 in the above equation:
 $ \Rightarrow \left( {{a_m} + {a_n}} \right) = - \left( {m + n - 1} \right)d + \left( {m + n - 1} \right)d$
Solving the right side, we get:
$ \Rightarrow \left( {{a_m} + {a_n}} \right) = 0$
Therefore, $\left( {{a_m} + {a_n}} \right)th$ term is zero.

Note: AP stands for Arithmetic series, it is a series in which the difference of two continuous terms is always constant, and the constant value is known as the common difference.
Remember to solve the equation step by step rather than substituting the value at once.