Let first term of A.P. be a and common difference be d. Hence we can write, \[ m[ a + (m-1)d ] = n[ a + (n-1)d ] \] \[ \therefore am + md(m-1) = an + nd(n-1) \] \[ \therefore am-an +md(m-1) - nd(n-1) =0 \] \[\therefore a(m-n) + d( m(m-1) - n(n-1) ) = 0 \] \[\therefore a(m-n) + d ( m^2 - n^2 -m + n ) = 0 \] \[\therefore a(m-n) + d[ (m-n)(m+n) - (m-n) ] =0 \cdots (i) \] As we know that \( m \neq n\) , so we can write that \( m-n \neq 0 \) Hence, we can divide the above equation \( (i) \) by \( m-n\) then we get, \[ a + [ (m+n) - 1 ]d = 0 \cdots (ii) \] Now the \( (m+n)^{th} \) of the A.P. is equal to \[ a + [(m+n) -1 ]d \cdots (iii) \]
From \( (ii) and (iii) \) we get ,\[ a + [(m+n) -1] d=0 \] i.e. \( (m+n)^{th} \) term of A.P. is \( 0\)
Hence, proved.
–
Kunal Joshi
·
2 years, 3 months ago

Let \(x_a\) represent the \(a^{th}\) term of the progression. Since this is an arithmetic progression, there exist \(c,d \in \mathbb{R}\) such that \(x_a = c*a + d\) for all \(a \in \mathbb{N}\). Then,
\[m*x_m = n*x_n\]
\[m(cm+d) = n(cn+d)\]
\[cm^2+dm = cn^2 + dn\]
\[c(m^2-n^2)+d(m-n)=0\]

Since \(m \neq n\), we can divide by \(m - n\):

\[c(m+n) + d = 0\]

Therefore, the \((m+n)^{th}\) term is 0.
–
Ariel Gershon
·
2 years, 3 months ago

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@Ariel Gershon
–
@ariel gershon what do c denotes here cos in india we have a different way to represent any term i.e. a+(n-1)*d where a is the first term n is the no. of term to find and d is the difference
–
Palash Som
·
2 years, 3 months ago

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@Palash Som
–
\(c\) and \(d\) are just constants. I agree, your formula is more commonly used, but you can rewrite it as \(d*n + (a-d)\), which fits the formula I used. I realize now that my method isn't that different from Kunal's; but I posted my solution before seeing his.
–
Ariel Gershon
·
2 years, 3 months ago

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TopNewestLet first term of A.P. be

aand common difference bed. Hence we can write, \[ m[ a + (m-1)d ] = n[ a + (n-1)d ] \] \[ \therefore am + md(m-1) = an + nd(n-1) \] \[ \therefore am-an +md(m-1) - nd(n-1) =0 \] \[\therefore a(m-n) + d( m(m-1) - n(n-1) ) = 0 \] \[\therefore a(m-n) + d ( m^2 - n^2 -m + n ) = 0 \] \[\therefore a(m-n) + d[ (m-n)(m+n) - (m-n) ] =0 \cdots (i) \] As we know that \( m \neq n\) , so we can write that \( m-n \neq 0 \) Hence, we can divide the above equation \( (i) \) by \( m-n\) then we get, \[ a + [ (m+n) - 1 ]d = 0 \cdots (ii) \] Now the \( (m+n)^{th} \) of the A.P. is equal to \[ a + [(m+n) -1 ]d \cdots (iii) \] From \( (ii) and (iii) \) we get ,\[ a + [(m+n) -1] d=0 \] i.e. \( (m+n)^{th} \) term of A.P. is \( 0\) Hence, proved. – Kunal Joshi · 2 years, 3 months agoLog in to reply

– Murtuza Akhtari · 2 years, 3 months ago

nice answer Kunal but can u please elaborate the stepsLog in to reply

– Palash Som · 2 years, 3 months ago

nicely done thnx..Log in to reply

Let \(x_a\) represent the \(a^{th}\) term of the progression. Since this is an arithmetic progression, there exist \(c,d \in \mathbb{R}\) such that \(x_a = c*a + d\) for all \(a \in \mathbb{N}\). Then, \[m*x_m = n*x_n\] \[m(cm+d) = n(cn+d)\] \[cm^2+dm = cn^2 + dn\] \[c(m^2-n^2)+d(m-n)=0\]

Since \(m \neq n\), we can divide by \(m - n\):

\[c(m+n) + d = 0\]

Therefore, the \((m+n)^{th}\) term is 0. – Ariel Gershon · 2 years, 3 months ago

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– Palash Som · 2 years, 3 months ago

@ariel gershon what do c denotes here cos in india we have a different way to represent any term i.e. a+(n-1)*d where a is the first term n is the no. of term to find and d is the differenceLog in to reply

– Ariel Gershon · 2 years, 3 months ago

\(c\) and \(d\) are just constants. I agree, your formula is more commonly used, but you can rewrite it as \(d*n + (a-d)\), which fits the formula I used. I realize now that my method isn't that different from Kunal's; but I posted my solution before seeing his.Log in to reply

– Palash Som · 2 years, 3 months ago

thnx a lot .. :-)Log in to reply

Bro dis sum is of 10th ssc hot sums ._. Don't copy :/ – Vedant Gaikwad · 2 years, 3 months ago

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– Palash Som · 2 years, 2 months ago

hey @vedant this one is a question in a book RS AGGARWALLog in to reply