Answer
Verified
428.1k+ views
Hint- In an A.P \[{{\text{n}}^{th}}\] Term is given as \[a + \left( {n - 1} \right)d\] where \[a\] is the first term and \[d\] is the common difference of an A.P.
In the question above it is given that \[{{\text{p}}^{th}}\]term of an A.P. is \[q\] and \[{{\text{q}}^{th}}\] term is \[p\] of an A.P.
For the given question \[{{\text{n}}^{th}}\] Term of an A.P is asked, to find it we know in general form \[{{\text{n}}^{th}}\] Term is given as \[a + \left( {n - 1} \right)d\] where \[a\] is the first term and \[d\] is the common difference of an A.P.
So to solve this question first let us assume \[a\] be the first term and \[d\] is the common difference of the given Arithmetic progression.
So we can write \[{{\text{p}}^{th}}\]term and \[{{\text{q}}^{th}}\] term of an A.P as
\[{{\text{p}}^{th}}{\text{ term }} = q \Rightarrow a + \left( {p - 1} \right)d = q{\text{ }}........\left( 1 \right)\]
And similarly
\[{{\text{q}}^{th}}{\text{ term }} = p \Rightarrow a + \left( {q - 1} \right)d = p{\text{ }}........\left( 2 \right)\]
From the above two equations we can find the value of $a$ and $d$ which we need to find the \[{{\text{n}}^{th}}\] Term.
So, we will subtract equation (2) from (1), from here we will get $d$
\[\left( {p - q} \right)d = \left( {q - p} \right) \Rightarrow d = - 1\]
And now the value of \[d\]obtained above we will put in equation (1), from here we will get $a$ value
\[{\text{i}}{\text{.e }}a + \left( {p - 1} \right) \times \left( { - 1} \right) = q \Rightarrow a = \left( {p + q - 1} \right)\]
So we need to find the \[{{\text{n}}^{th}}\] Term
\[{{\text{n}}^{th}}\] Term \[ = a + \left( {n - 1} \right)d = \left( {p + q - 1} \right) + \left( {n - 1} \right) \times - 1 = \left( {p + q - n} \right)\]
Hence Proved the \[{{\text{n}}^{th}}\] term is \[\left( {p + q - n} \right).\]
Note- Whenever this type of question appears it is important to note down given details as in this question it is given \[{{\text{p}}^{th}}\]term of an A.P. is \[q\] and \[{{\text{q}}^{th}}\] term is \[p\]. In Arithmetic Progression the difference between the two successive terms is same and we call it common difference \[d\].In an A.P \[{{\text{n}}^{th}}\] Term is given as \[a + \left( {n - 1} \right)d\] where \[a\] is the first term and \[d\] is the common difference of an A.P. Approach this type of question with intent to find the value of \[a\]and \[d\].
In the question above it is given that \[{{\text{p}}^{th}}\]term of an A.P. is \[q\] and \[{{\text{q}}^{th}}\] term is \[p\] of an A.P.
For the given question \[{{\text{n}}^{th}}\] Term of an A.P is asked, to find it we know in general form \[{{\text{n}}^{th}}\] Term is given as \[a + \left( {n - 1} \right)d\] where \[a\] is the first term and \[d\] is the common difference of an A.P.
So to solve this question first let us assume \[a\] be the first term and \[d\] is the common difference of the given Arithmetic progression.
So we can write \[{{\text{p}}^{th}}\]term and \[{{\text{q}}^{th}}\] term of an A.P as
\[{{\text{p}}^{th}}{\text{ term }} = q \Rightarrow a + \left( {p - 1} \right)d = q{\text{ }}........\left( 1 \right)\]
And similarly
\[{{\text{q}}^{th}}{\text{ term }} = p \Rightarrow a + \left( {q - 1} \right)d = p{\text{ }}........\left( 2 \right)\]
From the above two equations we can find the value of $a$ and $d$ which we need to find the \[{{\text{n}}^{th}}\] Term.
So, we will subtract equation (2) from (1), from here we will get $d$
\[\left( {p - q} \right)d = \left( {q - p} \right) \Rightarrow d = - 1\]
And now the value of \[d\]obtained above we will put in equation (1), from here we will get $a$ value
\[{\text{i}}{\text{.e }}a + \left( {p - 1} \right) \times \left( { - 1} \right) = q \Rightarrow a = \left( {p + q - 1} \right)\]
So we need to find the \[{{\text{n}}^{th}}\] Term
\[{{\text{n}}^{th}}\] Term \[ = a + \left( {n - 1} \right)d = \left( {p + q - 1} \right) + \left( {n - 1} \right) \times - 1 = \left( {p + q - n} \right)\]
Hence Proved the \[{{\text{n}}^{th}}\] term is \[\left( {p + q - n} \right).\]
Note- Whenever this type of question appears it is important to note down given details as in this question it is given \[{{\text{p}}^{th}}\]term of an A.P. is \[q\] and \[{{\text{q}}^{th}}\] term is \[p\]. In Arithmetic Progression the difference between the two successive terms is same and we call it common difference \[d\].In an A.P \[{{\text{n}}^{th}}\] Term is given as \[a + \left( {n - 1} \right)d\] where \[a\] is the first term and \[d\] is the common difference of an A.P. Approach this type of question with intent to find the value of \[a\]and \[d\].
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Summary of the poem Where the Mind is Without Fear class 8 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Write an application to the principal requesting five class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE