Question

Hola Brainlians. Here's a question to check your reasoning and Arithmetic Ability. The qúestion is from Arithmetic Progression .

If the sum of first m terms of an AP is equal to n and the sum of first n terms of that same AP is equal to m . Then show that the sum of first (m + n) terms of an AP is equal to -(m + n) .

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Answer 1

★ Concept :-

Here the concept of Arithmetic Progression has been used. We see that we are given sum relations that is sum of first m terms is equal to n and sum of first n terms is equal to m. So firstly according to formula we can equate the values in the relationships. Then from those both we can derive a new value which will help us to give the final answer.

Let's do it !!

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★ Formula Used :-

$\;\boxed{\sf{\pink{S_{p}\;=\;\bf{\dfrac{p}{2}\:[2a\:+\:(p\:-\:1)d]}}}}$

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★ Solution :-

Given,

» Sum of first m terms of an AP = Sm = n

» Sum of first n terms of an AP = Sn = m

• Let the first term of the AP be a

• Let the common difference of the AP be d

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By formula we know that,

$\;\sf{\rightarrow\;\;S_{p}\;=\;\bf{\dfrac{p}{2}\:[2a\:+\:(p\:-\:1)d]}}$

This formula shows the sum of first p terms of that AP.

• For the Sum of First m terms ::

Now replacing p with m we get,

$\;\sf{\rightarrow\;\;S_{m}\;=\;\bf{\dfrac{m}{2}\:[2a\:+\:(m\:-\:1)d]}}$

Now applying the value of Sm we get,

$\;\sf{\rightarrow\;\;n\;=\;\bf{\dfrac{m}{2}\:[2a\:+\:(m\:-\:1)d]}}$

$\;\sf{\rightarrow\;\;\dfrac{m}{2}\:[2a\:+\:(m\:-\:1)d]\;=\;\bf{n}}$

$\;\sf{\rightarrow\;\;m[2a\:+\:(m\:-\:1)d]\;=\;\bf{2n}}$

$\;\sf{\rightarrow\;\;2am\:+\:m(dm\:-\:d)\;=\;\bf{2n}}$

$\;\sf{\rightarrow\;\;\green{2am\:+\:dm^{2}\:-\:dm\;=\;\bf{2n}}}$

Let this be equation (i) .

• For the Sum of First n terms ::

Now replacing p with n in that same formula, we get

$\;\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{n}{2}\:[2a\:+\:(n\:-\:1)d]}}$

Now by applying the value of Sn we get,

$\;\sf{\rightarrow\;\;m\;=\;\bf{\dfrac{n}{2}\:[2a\:+\:(n\:-\:1)d]}}$

$\;\sf{\rightarrow\;\;\dfrac{n}{2}\:[2a\:+\:(n\:-\:1)d]\;=\;\bf{m}}$

$\;\sf{\rightarrow\;\;n[2a\:+\:(n\:-\:1)d]\;=\;\bf{2m}}$

$\;\sf{\rightarrow\;\;2an\:+\:n(dn\:-\:d)\;=\;\bf{2m}}$

$\;\sf{\rightarrow\;\;\blue{2an\:+\:dn^{2}\:-\:dn\;=\;\bf{2m}}}$

Let this be equation (ii).

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Now by subtracting equation (ii) from equation (i) we get,

$\;\sf{\rightarrow\;\;(2am\:+\:dm^{2}\:-\:dm)\:-\:(2an\:+\:dn^{2}\:-\:dn)\;=\;\bf{2n\:-\:2m}}$

$\;\sf{\rightarrow\;\;2am\:+\:dm^{2}\:-\:dm\:-\:2an\:-\:dn^{2}\:+\:dn\;=\;\bf{2n\:-\:2m}}$

$\;\sf{\rightarrow\;\;2am\:+\:dm^{2}\:-\:dm\:-\:2an\:-\:dn^{2}\:+\:dn\;=\;\bf{2n\:-\:2m}}$

$\;\sf{\rightarrow\;\;2am\:+\:dm^{2}\:-\:dm\:-\:2an\:-\:dn^{2}\:+\:dn\;=\;\bf{2n\:-\:2m}}$

On rearranging the terms we get,

$\;\sf{\rightarrow\;\;2am\:-\:2an\:+\:dm^{2}\:-\:dn^{2}\:+\:dn\:-\:dm\;=\;\bf{2n\:-\:2m}}$

$\;\sf{\rightarrow\;\;2a(m\:-\:n)\:+\:[dm^{2}\:-\:dn^{2}\:+\:dn\:-\:dm]\;=\;\bf{2(n\:-\:m)}}$

Now taking d as common from the terms in bracket, we get

$\;\sf{\rightarrow\;\;2a(m\:-\:n)\:+\:[(m^{2}\:-\:n^{2})\:+\:(n\:-\:m)]d\;=\;\bf{2(n\:-\:m)}}$

This can be written as ,

$\;\sf{\rightarrow\;\;2a(m\:-\:n)\:+\:[(m^{2}\:-\:n^{2})\:-\:(m\:-\:n)]d\;=\;\bf{-2(m\:-\:n)}}$

By identity, we know that :-

→ x² - y² = (x + y)(x - y)

• Here x² = m²

• Here y² = n²

By applying these we get,

$\;\sf{\rightarrow\;\;2a(m\:-\:n)\:+\:[(m\:-\:n)(m\:+\:n)\:-\:(m\:-\:n)]d\;=\;\bf{-2(m\:-\:n)}}$

Now taking (m - n) term as common from LHS, we get

$\;\sf{\rightarrow\;\;(m\:-\:n)\bigg(2a\:+\:[(m\:+\:n)\:-\:1]d\bigg)\;=\;\bf{-2(m\:-\:n)}}$

$\;\sf{\rightarrow\;\;(m\:-\:n)[2a\:+\:(m\:+\:n\:-\:1)d]\;=\;\bf{-2(m\:-\:n)}}$

Now cancelling (m - n) term from both sides, we get

$\;\sf{\rightarrow\;\;\red{2a\:+\:(m\:+\:n\:-\:1)d\;=\;\bf{-2}}}$

Let this be equation (iii).

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Now replacing p with (m + n) in the main formula of sum of first p terms , we get

$\;\sf{\rightarrow\;\;S_{m\:+\:n}\;=\;\bf{\dfrac{m\:+\:n}{2}\:[2a\:+\:(m\:+\:n\:-\:1)d]}}$

This gives ths sum of first (m + n) terms.

Now from equation iii), by applying the required value, we get

$\;\sf{\rightarrow\;\;S_{m\:+\:n}\;=\;\bf{\dfrac{m\:+\:n}{2}\:[-2]}}$

By cancelling 2 , we get

$\;\sf{\rightarrow\;\;S_{m\:+\:n}\;=\;\bf{\dfrac{m\:+\:n}{\not{2}}\:[-\not{2}]}}$

$\;\sf{\rightarrow\;\;S_{m\:+\:n}\;=\;\bf{(m\:+\:n)\:\times\:-1}}$

$\;\large{\bf{\rightarrow\;\;\purple{S_{m\:+\:n}\;=\;\bf{-(m\:+\:n)}}}}$

Hence proved.

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★ More to know :-

$\;\tt{a_{n}\;=\;a\;+\;(n\:-\:1)d}$

$\;\tt{S_{n}\;=\;\dfrac{n}{2}[a\:+\:a_{n}]}$

$\;\tt{S_{n}\;=\;\dfrac{n}{2}[a\;-\;l_{n}]}$

Where ln is the last term of the A.P.

Answer 2

$\huge \sf \red{Question}$

If the sum of first m terms of an AP is equal to n and the sum of first n terms of that same AP is equal to m . Then show that the sum of first (m + n) terms of an AP is equal to -(m + n)

$\huge\sf \red{Answer}$

$\sf{Let  a  be \: the \: first \: term \: and \:  d  \: be \: the \: common \: difference \: of \: the \: given \: AP. Then,</p><p></p><p></p><p>}$

$\sf sn \: \frac{n}{2}[2a+(n−1)d]=m$

Given,

Sm=n

$\sf{Sn= \frac{m}{2}[2a+(n−1)d]}$

$\sf{Sn=m}$

$\sf \frac{n}{2}[2a+(n−1)d]=m$

$\sf{2an+n(n−1)d=2m          ...(2)}$

On subtracting 2 from 1 , we get,

$\sf{2a(m−n)+[(m²−n²)−(m−n)]d=2(n−m)}$

$\sf{(m−n)[2a+(m+n−1)d]=2(n−m)}$

$\sf{2a+(m+n−1)d=−2          ...(3)}$

$\sf{Sum \: of (m+n) terms \: of \: the \: given \: AP}$

$\sf{Sm+n= \frac{m+n}{2} \: [2a+(m+n−1)d]}$

$\sf{= \frac{m+n}{2}(−2)=−(m+n) }$

$\sf{Hence \: proved.}$