Answer
Verified
426.9k+ views
Hint: Whenever it is given that the terms are in A.P, try to use the formulas of A.P, it can be the formula to find the first term or the last term or the sum which makes the question easier to solve as we obtain some equations by putting the values from the given question.
Let $a$ be the first term and $d$ be the common difference of the given A.P.
Now to find the sum we are going to find the general form of m terms and n terms.
${a_m} = \dfrac{1}{n} \Rightarrow a + \left( {m - 1} \right)d = \dfrac{1}{n}$ $......\left( 1 \right)$
\[{a_n} = \dfrac{1}{m} \Rightarrow a + \left( {n - 1} \right)d = \dfrac{1}{m}\] \[......\left( 2 \right)\]
Now we are going to subtract \[\left( 2 \right)\] from\[\left( 1 \right)\], we get,
\[\left( {m - n} \right)d = \dfrac{1}{n} - \dfrac{1}{m} \Rightarrow \left( {m - n} \right)d = \dfrac{{m - n}}{{mn}} \Rightarrow d = \dfrac{1}{{mn}}\]
The next step is to equate \[d = \dfrac{1}{{mn}}\] in\[\left( 1 \right)\], the equation becomes,
\[a + \left( {m - 1} \right)\dfrac{1}{{mn}} = \dfrac{1}{n} \Rightarrow a + \dfrac{1}{n} - \dfrac{1}{{mn}} = \dfrac{1}{n} \Rightarrow a = \dfrac{1}{{mn}}\]
To find the sum of an A.P of \[r\] terms with \[a\]as the first term and \[d\] as the common difference the formula is:
\[{S_r} = \dfrac{r}{2}\left\{ {2a + \left( {n - 1} \right)d} \right\}\]
Therefore, if we substitute the value of $a$and $d$ in the formula of sum of A.P., we get,
\[{S_{mn}} = \dfrac{{mn}}{2}\left\{ {2a + \left( {mn - 1} \right)d} \right\}\]
Note: In this question, they have asked us to find the sum of total terms, therefore we are going to take\[r\] as $mn$ here.
In the next step we equate the value of,
\[{S_{mn}} = \dfrac{{mn}}{2}\left\{ {\dfrac{2}{{mn}} + \left( {mn - 1} \right) \times \dfrac{1}{{mn}}} \right\} = \dfrac{1}{2}\left( {mn + 1} \right)\]
Hence Proved!
Let $a$ be the first term and $d$ be the common difference of the given A.P.
Now to find the sum we are going to find the general form of m terms and n terms.
${a_m} = \dfrac{1}{n} \Rightarrow a + \left( {m - 1} \right)d = \dfrac{1}{n}$ $......\left( 1 \right)$
\[{a_n} = \dfrac{1}{m} \Rightarrow a + \left( {n - 1} \right)d = \dfrac{1}{m}\] \[......\left( 2 \right)\]
Now we are going to subtract \[\left( 2 \right)\] from\[\left( 1 \right)\], we get,
\[\left( {m - n} \right)d = \dfrac{1}{n} - \dfrac{1}{m} \Rightarrow \left( {m - n} \right)d = \dfrac{{m - n}}{{mn}} \Rightarrow d = \dfrac{1}{{mn}}\]
The next step is to equate \[d = \dfrac{1}{{mn}}\] in\[\left( 1 \right)\], the equation becomes,
\[a + \left( {m - 1} \right)\dfrac{1}{{mn}} = \dfrac{1}{n} \Rightarrow a + \dfrac{1}{n} - \dfrac{1}{{mn}} = \dfrac{1}{n} \Rightarrow a = \dfrac{1}{{mn}}\]
To find the sum of an A.P of \[r\] terms with \[a\]as the first term and \[d\] as the common difference the formula is:
\[{S_r} = \dfrac{r}{2}\left\{ {2a + \left( {n - 1} \right)d} \right\}\]
Therefore, if we substitute the value of $a$and $d$ in the formula of sum of A.P., we get,
\[{S_{mn}} = \dfrac{{mn}}{2}\left\{ {2a + \left( {mn - 1} \right)d} \right\}\]
Note: In this question, they have asked us to find the sum of total terms, therefore we are going to take\[r\] as $mn$ here.
In the next step we equate the value of,
\[{S_{mn}} = \dfrac{{mn}}{2}\left\{ {\dfrac{2}{{mn}} + \left( {mn - 1} \right) \times \dfrac{1}{{mn}}} \right\} = \dfrac{1}{2}\left( {mn + 1} \right)\]
Hence Proved!
Recently Updated Pages
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Find the values of other five trigonometric functions class 10 maths CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Discuss what these phrases mean to you A a yellow wood class 9 english CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Our body works within the pH range of A 665 B 778 C class 9 chemistry CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Name 10 Living and Non living things class 9 biology CBSE
Write an Article on Save Earth Save Life