# PROVE IT !!!

if m times the m term of an A.P. is equal to the n times the n term and (m) is not equal to (n) show that (m+n) term is 0 .

Note by Palash Som
3 years, 9 months ago

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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.

- 3 years, 9 months ago

- 3 years, 9 months ago

nicely done thnx..

- 3 years, 9 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.

- 3 years, 9 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 difference

- 3 years, 9 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.

- 3 years, 9 months ago

thnx a lot .. :-)

- 3 years, 9 months ago

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

- 3 years, 9 months ago

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

- 3 years, 8 months ago