# In an A.P., if ${{\text{m}}^{{\text{th}}}}$term is n and the ${{\text{n}}^{{\text{th}}}}$ term is m, where m$ \ne $n find the ${{\text{p}}^{{\text{th}}}}$ term.

A . m+n-p

B . m−n+p

C . m+n−p

D . m−n−p

Last updated date: 24th Mar 2023

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Answer

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Hint:In order to solve to this problem, use general ${{\text{n}}^{{\text{th}}}}$term of different arithmetic progressions(Aps) which is ${{\text{a}}_{\text{n}}}$= a + (n−1)d so that we can find first and last term of AP which is required to get ${{\text{p}}^{{\text{th}}}}$ term.

Complete step-by-step answer:

We have the ${{\text{n}}^{{\text{th}}}}$term of an AP,

For

${{\text{a}}_{\text{n}}}$=a+(n−1) d

Where, a is the${{\text{n}}^{{\text{th}}}}$ term first term and d is the common difference

As we find ${{\text{n}}^{{\text{th}}}}$ term similarly we can find ${{\text{m}}^{{\text{th}}}}$ term

For ${{\text{m}}^{{\text{th}}}}$ term,

${{\text{a}}_{\text{m}}}$=a + (m−1) d

Where, a is the first term and d is the common difference

In the question it is given that ${{\text{m}}^{{\text{th}}}}$term ${{\text{a}}_{\text{m}}}$is equal to n

${{\text{a}}_{\text{m}}}$=a + (m−1)d = n ................(1)

In the question it is given that ${{\text{n}}^{{\text{th}}}}$term ${{\text{a}}_{\text{n}}}$is equal to m

an=a + (n−1)d = m ..............(2)

on subtracting equation (2) from equation (1),

a+(m−1)d−(a+(n−1)d ) = n−m

(m−1)d−(n-1) d = n−m

On further solving

(m−1−n+1) d = n−m

(m−n)d = n−m

$ \Rightarrow {\text{d = }}\dfrac{{{\text{n - m}}}}{{{\text{m - n}}}}$

$ \Rightarrow {\text{d = - 1}}$ ; here we get common difference of AP

Substitute d = -1 in equation (1),

a+(m−1) (−1) = n

a−m+1=n

a = n+m−1; Here we get first term of AP

For ${{\text{p}}^{{\text{th}}}}$ term

${{\text{a}}_{\text{p}}}$=a+(p−1)d

On putting a= n+m−1 & d = -1 in above equation

= n+m−1+(p−1) (−1)

=n+m−1−p+1

= n+m−p ; Which is required ${{\text{p}}^{{\text{th}}}}$term

Hence Option C is correct.

Note: Whenever we face such type of problems we must choose expressions of nth term along with the proper understanding of common difference and first term of different APs, because by applying further proper mathematics like Addition or subtraction on general nth term we can get our desired result.

Complete step-by-step answer:

We have the ${{\text{n}}^{{\text{th}}}}$term of an AP,

For

${{\text{a}}_{\text{n}}}$=a+(n−1) d

Where, a is the${{\text{n}}^{{\text{th}}}}$ term first term and d is the common difference

As we find ${{\text{n}}^{{\text{th}}}}$ term similarly we can find ${{\text{m}}^{{\text{th}}}}$ term

For ${{\text{m}}^{{\text{th}}}}$ term,

${{\text{a}}_{\text{m}}}$=a + (m−1) d

Where, a is the first term and d is the common difference

In the question it is given that ${{\text{m}}^{{\text{th}}}}$term ${{\text{a}}_{\text{m}}}$is equal to n

${{\text{a}}_{\text{m}}}$=a + (m−1)d = n ................(1)

In the question it is given that ${{\text{n}}^{{\text{th}}}}$term ${{\text{a}}_{\text{n}}}$is equal to m

an=a + (n−1)d = m ..............(2)

on subtracting equation (2) from equation (1),

a+(m−1)d−(a+(n−1)d ) = n−m

(m−1)d−(n-1) d = n−m

On further solving

(m−1−n+1) d = n−m

(m−n)d = n−m

$ \Rightarrow {\text{d = }}\dfrac{{{\text{n - m}}}}{{{\text{m - n}}}}$

$ \Rightarrow {\text{d = - 1}}$ ; here we get common difference of AP

Substitute d = -1 in equation (1),

a+(m−1) (−1) = n

a−m+1=n

a = n+m−1; Here we get first term of AP

For ${{\text{p}}^{{\text{th}}}}$ term

${{\text{a}}_{\text{p}}}$=a+(p−1)d

On putting a= n+m−1 & d = -1 in above equation

= n+m−1+(p−1) (−1)

=n+m−1−p+1

= n+m−p ; Which is required ${{\text{p}}^{{\text{th}}}}$term

Hence Option C is correct.

Note: Whenever we face such type of problems we must choose expressions of nth term along with the proper understanding of common difference and first term of different APs, because by applying further proper mathematics like Addition or subtraction on general nth term we can get our desired result.

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