# If the \[{{\text{p}}^{th}}\]term of an A.P. is \[q\] and \[{{\text{q}}^{th}}\] term is \[p\], prove that its \[{{\text{n}}^{th}}\] term is \[\left( {p + q - n} \right).\]

Last updated date: 24th Mar 2023

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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 \[d\].

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