Answer

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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..,${t_3} - {t_2} = {t_2} - {t_1} = {t_n} - {t_{n - 1}} = d$.

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..,${t_3} - {t_2} = {t_2} - {t_1} = {t_n} - {t_{n - 1}} = d$.

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