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Question: If the \({m^{th}}\) term of an A.P be \(\dfrac{1}{n}\) and \({n^{th}}\) term be \(\dfrac{1}{m}\) , t...

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

Explanation

Solution

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

Complete step-by-step answer:
Given,
The mth{m^{th}}term of an A.P is 1n\dfrac{1}{n} and nth{n^{th}}term is 1m\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 nth{n^{th}}term of an A.P will be Tn=a+(n−1)d{T_n} = a + (n - 1)d.
Therefore, the mth{m^{th}} term can be written as
⇒mthterm=1n ⇒1n=a+(m−1)d→(1)  \Rightarrow {m^{th}}term = \dfrac{1}{n} \\\ \Rightarrow \dfrac{1}{n} = a + (m - 1)d \to (1) \\\
Similarly, the nth{n^{th}}term is given as 1m\dfrac{1}{m}, it can be written as
⇒nthterm=1m ⇒1m=a+(n−1)d→(2)  \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
⇒1n−1m=a+(m−1)d−a−(n−1)d ⇒1n−1m=md−d−nd+d ⇒m−nmn=(m−n)d ⇒1mn=d  \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.., 1mn\dfrac{1}{{mn}}.Now let us substitute the obtained value of d in equation (1), we get
(1)⇒1n=a+(m−1)d ⇒1n=a+(m−1)mn ⇒1n=a+1n−1mn ⇒1n−1n+1mn=a ⇒1mn=a  (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 1mn\dfrac{1}{{mn}}.Therefore, we can say
a=d=1mna = d = \dfrac{1}{{mn}}
Now, we need to find the (mn)th{(mn)^{th}}term which is equal to ‘a+(mn−1)da + (mn - 1)d’.So substituting the obtained values of ‘a’ and d’ we get
⇒(mn)thterm=a+(mn−1)d=1mn+(mn−1)(1mn)=1mn+1−1mn=1 ⇒(mn)thterm=1  \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{(mn)^{th}}term is11.

Note: In an AP, d is the common difference of the consecutive terms i.e..,t3−t2=t2−t1=tn−tn−1=d{t_3} - {t_2} = {t_2} - {t_1} = {t_n} - {t_{n - 1}} = d.