# 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 \[\left( {p + q - n} \right).\]

Answer

Verified

364.8k+ 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 \[d\].

Last updated date: 25th Sep 2023

â€¢

Total views: 364.8k

â€¢

Views today: 7.64k

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

Difference Between Plant Cell and Animal Cell

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

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Drive an expression for the electric field due to an class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

What is the past tense of read class 10 english CBSE