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.

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$

@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

$c$ 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.

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## Comments

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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.Log in to reply

nice answer Kunal but can u please elaborate the steps

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nicely done thnx..

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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.

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@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

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$c$ 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.

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Bro dis sum is of 10th ssc hot sums ._. Don't copy :/

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hey @vedant this one is a question in a book RS AGGARWAL

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