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
361.8k+ views
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..,${t_3} - {t_2} = {t_2} - {t_1} = {t_n} - {t_{n - 1}} = d$.
Last updated date: 24th Sep 2023
•
Total views: 361.8k
•
Views today: 4.61k
Recently Updated Pages
What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

Difference Between Plant Cell and Animal Cell

What is the basic unit of classification class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

One cusec is equal to how many liters class 8 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers
