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
5 years, 1 month ago

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Let first term of A.P. be a and common difference be d. Hence we can write, m[a+(m1)d]=n[a+(n1)d] m[ a + (m-1)d ] = n[ a + (n-1)d ] am+md(m1)=an+nd(n1) \therefore am + md(m-1) = an + nd(n-1) aman+md(m1)nd(n1)=0 \therefore am-an +md(m-1) - nd(n-1) =0 a(mn)+d(m(m1)n(n1))=0\therefore a(m-n) + d( m(m-1) - n(n-1) ) = 0 a(mn)+d(m2n2m+n)=0\therefore a(m-n) + d ( m^2 - n^2 -m + n ) = 0 a(mn)+d[(mn)(m+n)(mn)]=0(i)\therefore a(m-n) + d[ (m-n)(m+n) - (m-n) ] =0 \cdots (i) As we know that mn m \neq n , so we can write that mn0 m-n \neq 0 Hence, we can divide the above equation (i) (i) by mn m-n then we get, a+[(m+n)1]d=0(ii) a + [ (m+n) - 1 ]d = 0 \cdots (ii) Now the (m+n)th (m+n)^{th} of the A.P. is equal to a+[(m+n)1]d(iii) a + [(m+n) -1 ]d \cdots (iii) From (ii)and(iii) (ii) and (iii) we get ,a+[(m+n)1]d=0 a + [(m+n) -1] d=0 i.e. (m+n)th (m+n)^{th} term of A.P. is 0 0 Hence, proved.

Kunal Joshi - 5 years, 1 month ago

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nice answer Kunal but can u please elaborate the steps

Murtuza Akhtari - 5 years, 1 month ago

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nicely done thnx..

Palash Som - 5 years, 1 month ago

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Let xax_a represent the atha^{th} term of the progression. Since this is an arithmetic progression, there exist c,dRc,d \in \mathbb{R} such that xa=ca+dx_a = c*a + d for all aNa \in \mathbb{N}. Then, mxm=nxnm*x_m = n*x_n m(cm+d)=n(cn+d)m(cm+d) = n(cn+d) cm2+dm=cn2+dncm^2+dm = cn^2 + dn c(m2n2)+d(mn)=0c(m^2-n^2)+d(m-n)=0

Since mnm \neq n, we can divide by mnm - n:

c(m+n)+d=0c(m+n) + d = 0

Therefore, the (m+n)th(m+n)^{th} term is 0.

Ariel Gershon - 5 years, 1 month ago

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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

Palash Som - 5 years, 1 month ago

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cc and dd are just constants. I agree, your formula is more commonly used, but you can rewrite it as dn+(ad)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 - 5 years, 1 month ago

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@Ariel Gershon thnx a lot .. :-)

Palash Som - 5 years, 1 month ago

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

Vedant Gaikwad - 5 years, 1 month ago

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

Palash Som - 4 years, 12 months ago

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