Question

If $m$ times the ${m^{th}}$ term of an A.P. is equal to $n$ times its ${n^{th}}$ term, find the ${(m + n)^{th}}$ term of the A.P.
$
  {\text{A}}{\text{. 1}} \\
  {\text{B}}{\text{. 0}} \\
  {\text{C}}{\text{. - 1}} \\
  {\text{D}}{\text{. 2}} \\
$

Answer 1

Hint:- ${n^{th}}$term of an A.P. is given by $a + (n - 1)d$ where d is the common difference between two consecutive terms. Using this formula write the terms given in the question therefore equating and rearranging those terms we will get the ${(m + n)^{th}}$ term.

Complete step-by-step answer:
In this question
${n^{th}}$term of given A.P. =$a + (n - 1)d$ eq1.
${m^{th}}$term of given A.P. =$a + (m - 1)d$ eq2.
Now it is given that $m$ times the ${m^{th}}$ term of an A.P. is equal to $n$ times its ${n^{th}}$ term
Thus from eq1. , eq2.
$ \Rightarrow $$m\{ a + (m - 1)d\} = n\{ a + (n - 1)d\} $
On solving above equation
$
   \Rightarrow ma + (m - 1)md = na + (n - 1)nd \\
   \Rightarrow (m - n)a + ({m^2}^{^{}} - m)d - ({n^2} - n)d = 0 \\
   \Rightarrow (m - n)a + ({m^2} - {n^2})d - (m - n)d = 0 \\
    \\
 $
 On taking common from the above equation we get
$
   \Rightarrow (m - n)\{ a + (m + n)d - d\} = 0 \\
   \Rightarrow (m - n)\{ a + (m + n - 1)d\} = 0 \\
    \\
$
Since $m \ne n$ because ${m^{th}}$ and ${n^{th}}$ both terms are different
$ \Rightarrow a + (m + n - 1)d = 0$ eq 3.
Above equation represents the ${(m + n)^{th}}$ term of the given A.P.
Hence option B is correct.

Note:- Whenever you get this type of question the key concept of solving is you have to write the given terms of A.P. in standard form like $a + (n - 1)d$ and then compare them. And get the resultant equation after solving the formed equation.