Question

If the ${m^{th}}$ term of an A.P be $\dfrac{1}{n}$ and ${n^{th}}$ term be $\dfrac{1}{m}$ , then show that its ${(mn)^{th}}$ term is 1.

Answer 1

Hint: Here, we will use the Arithmetic Progression Concept and the ${n^{th}}$term formulae i.e..,${T_n} = a + (n - 1)d$.

Complete step-by-step answer:
Given,
The ${m^{th}}$term of an A.P is $\dfrac{1}{n}$ and ${n^{th}}$term is $\dfrac{1}{m}$
Now, let us consider ‘a’ be the first term and‘d’ is the common difference of an A.P.As, we know that the ${n^{th}}$term of an A.P will be ${T_n} = a + (n - 1)d$.
Therefore, the ${m^{th}}$ term can be written as
$
   \Rightarrow {m^{th}}term = \dfrac{1}{n} \\
   \Rightarrow \dfrac{1}{n} = a + (m - 1)d \to (1) \\
$
Similarly, the ${n^{th}}$term is given as $\dfrac{1}{m}$, it can be written as
$
   \Rightarrow {n^{th}}term = \dfrac{1}{m} \\
   \Rightarrow \dfrac{1}{m} = a + (n - 1)d \to (2) \\
 $
Let us subtract equation (2) from equation (1), we get
$
   \Rightarrow \dfrac{1}{n} - \dfrac{1}{m} = a + (m - 1)d - a - (n - 1)d \\
   \Rightarrow \dfrac{1}{n} - \dfrac{1}{m} = md - d - nd + d \\
   \Rightarrow \dfrac{{m - n}}{{mn}} = (m - n)d \\
   \Rightarrow \dfrac{1}{{mn}} = d \\
$
Hence, we obtained the value of d i.e.., $\dfrac{1}{{mn}}$.Now let us substitute the obtained value of d in equation (1), we get
$
  (1) \Rightarrow \dfrac{1}{n} = a + (m - 1)d \\
   \Rightarrow \dfrac{1}{n} = a + \dfrac{{(m - 1)}}{{mn}} \\
   \Rightarrow \dfrac{1}{n} = a + \dfrac{1}{n} - \dfrac{1}{{mn}} \\
   \Rightarrow \dfrac{1}{n} - \dfrac{1}{n} + \dfrac{1}{{mn}} = a \\
   \Rightarrow \dfrac{1}{{mn}} = a \\
$
Hence, here also we got the value of ‘a’ as $\dfrac{1}{{mn}}$.Therefore, we can say
$a = d = \dfrac{1}{{mn}}$
Now, we need to find the ${(mn)^{th}}$term which is equal to ‘$a + (mn - 1)d$’.So substituting the obtained values of ‘a’ and d’ we get
$
   \Rightarrow {(mn)^{th}}term = a + (mn - 1)d = \dfrac{1}{{mn}} + (mn - 1)(\dfrac{1}{{mn}}) = \dfrac{1}{{mn}} + 1 - \dfrac{1}{{mn}} = 1 \\
   \Rightarrow {(mn)^{th}}term = 1 \\
$
Hence, we proved that the value of ${(mn)^{th}}$term is$1$.

Note: In an AP, d is the common difference of the consecutive terms i.e..,${t_3} - {t_2} = {t_2} - {t_1} = {t_n} - {t_{n - 1}} = d$.