If m times the mth term of an A.P is equal to n times its nth term, show that the (m+n)th term of the A.P is zero.

Given :--

m{a +(m-1)d} = n{a +(n-1)d}

ma + m2d -md = na +n2d -nd

ma - na + m2d - n2d -md +nd = 0

a(m-n) +d(m2 - n2) -d(m-n) = 0

(m-n) (a + d(m+n) -d) = 0

 (a + d(m+n) -d) = 0

 a + (m+n - 1)d  = 0 

thus, (m+n)th term = 0....proved..

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Let mth term be am  and  nth term be an 

According to the question :

m(am ) = n(an )

Apply Formula : an = a + (n-1) *d  [ a - first term , d -Common Difference , n - No of terms ]

m( a + (m-1) *d ) = n( a + (n-1) *d )

{ma + (m2 - m) d }= [na + (n2 - n) d]

{ma + (m2 d- md) }= [na + (n2 d- nd) ]

[ma-na] + { m d-  n2 d } + (nd-md) = 0

a*[m-n] + d* { m -  n } + d* (n-m) = 0

Apply : a -  b2 = (a+b) (a-b)

a*[m-n] + d* { (m+n) (m-n) } - d* (m - n) = 0

Divide the above equation  with " (m-n)" ,We get :

a + d* { (m+n) } - d = 0

a + { (m+n) - 1 } d = 0

COMPARING THE EQUATION WITH : an = a + (n-1) *d 

Therefore : a + { (m+n) - 1 } d  = am+n 

So : am+n  = 0 

So  :the (m+n)th  term of the A.P is zero

  (HENCE PROVED)

If you are satisfied do give thumbs up.

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Given :-- m{a +(m-1)d} = n{a +(n-1)d} ma + m²d -md = na +n²d -nd ma - na + m²d - n²d -md +nd = 0 a(m-n) +d(m² - n²) -d(m-n) = 0 (m-n) (a + d(m+n) -d) = 0 (a + d(m+n) -d) = 0 Taking common (d) We get a + (m+n - 1)d = 0 thus, (m+n)th term = 0....proved.. For more you can contact on face book... Aman wadhwa
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Given m × tm = n × tn
⇒ m[a + (m – 1)d] = n[a + (n – 1)d]
⇒ ma + m(m – 1)d] = na + n(n – 1)d
⇒ ma + m(m – 1)d] – na – n(n – 1)d = 0
⇒ a(m – n) + [m2 – m – n2 + n]d = 0
⇒ a(m – n) + [m2 – n2 – m +  n]d = 0
⇒ a(m – n) + [(m – n)(m + n) – (m –  n)]d = 0
⇒ (m – n)[a+ {(m + n) – 1}]d = 0
⇒ [a+ {(m + n) – 1}]d = 0
Hence t(m + n) = 0

   
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Given m × tm = n × tn
⇒ m[a + (m – 1)d] = n[a + (n – 1)d]
⇒ ma + m(m – 1)d] = na + n(n – 1)d
⇒ ma + m(m – 1)d] – na – n(n – 1)d = 0
⇒ a(m – n) + [m2 – m – n2 + n]d = 0
⇒ a(m – n) + [m2 – n2 – m +  n]d = 0
⇒ a(m – n) + [(m – n)(m + n) – (m –  n)]d = 0
⇒ (m – n)[a+ {(m + n) – 1}]d = 0
⇒ [a+ {(m + n) – 1}]d = 0
Hence t(m + n) = 0

   
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Given m × tm = n × tn
⇒ m[a + (m – 1)d] = n[a + (n – 1)d]
⇒ ma + m(m – 1)d] = na + n(n – 1)d
⇒ ma + m(m – 1)d] – na – n(n – 1)d = 0
⇒ a(m – n) + [m2 – m – n2 + n]d = 0
⇒ a(m – n) + [m2 – n2 – m +  n]d = 0
⇒ a(m – n) + [(m – n)(m + n) – (m –  n)]d = 0
⇒ (m – n)[a+ {(m + n) – 1}]d = 0
⇒ [a+ {(m + n) – 1}]d = 0
Hence t(m + n) = 0
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Dear Student,   Thge general nth term of an AP is a + (n -1)d.   From the given conditions,   m (a + (m-1)d) = n( a + (n-1)d) => am + m2d - md = an + n2d - nd => a(m-n) + (m+n)(m-n)d - (m-n)d = 0 => (m-n) ( a + (m+n-1)d ) = 0   Rejecting the non-trivial case of m=n, we assume that m and n are different. => ( a + (m + n - 1)d ) = 0   The LHS of this equation denotes the (m+n)th term of the AP, which is Zero.    
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) Let the first term of the AP be 'a' and its common difference be 'd'. 

ii) So mth term is: {a + (m - 1)d} and nth term is [a + (n - 1)d} 

iii) As given, m{a + (m - 1) d} = n{a + (n - 1)d} 

Expanding, rearranging and simplifying, 

(m - n)a + (m² - n²)d - (m - n)d = 0 

Factorizing, (m - n)*[a + {(m + n) - 1}d] = 0 ---- (1) 

Since m & n are different the factor (m - n) is not equal to zero. 

Hence equation (1) = 0, implies that [a + {(m + n) - 1}d] = 0 

But this (m + n)th term of the AP. 

Thus it is proved that (m + n)th term of this AP = 0
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mtm = ntn  (Given)    m [a + (m – 1) d] = n [a + (n – 1) d]                                                                                                  [ tn = a + (n – 1) d]     m [a + md – d] = n [a + nd – d]     am + m2d – md = an + n2d – nd     am – an + m2d – n2d – md + nd = 0     a (m – n) + d (m2 – n2) – d (m – n) = 0     a (m – n) + d (m + n) (m – n) – d (m – n) = 0                                                                                                     [ a2 – b2 = (a + b) (a – b)]   Dividing through at by m – n, we get   a + d (m + n) - d = 0     a + d [m + n – 1] = 0     a + (m + n – 1) d = 0 ------------------------eq. no. (1)   tm+n = a + (m + n – 1)   [ tn = a + (n – 1) d]    tm + n = 0  [from eq. no. (1)]   Hence proved.  
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if m times the mth term is equal to n times the nth term of an ap. find its (m+n)th term
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if a(m+n)=0 then prove that (m+n)th term is 0
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Hi
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thank you so much
 
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thumbs up
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let mth term be a   and  nth term be a According to the question :
m(a  ) = n(a  ) Apply Formula : a = a + (n-1) *d  [ a - rst term , d -Common Difference , n - No of terms ]
m( a + (m-1) *d ) = n( a + (n-1) *d ) {ma + (m - m) d }= [na + (n - n) d] {ma + (m d- md) }= [na + (n d- nd) ] [ma-na] + { m d-  n d } + (nd-md) = 0 a*[m-n] + d* { m -  n  } + d* (n-m) = 0 Apply : a -  b = (a+b) (a-b) a*[m-n] + d* { (m+n) (m-n) } - d* (m - n) = 0 Divide the above equation  with " (m-n)" ,We get : a + d* { (m+n) } - d = 0 a + { (m+n) - 1 } d = 0 COMPARING THE EQUATION WITH : a = a + (n-1) *d Therefore : a + { (m+n) - 1 } d  = a So : a   = 0  So  :the (m+n)th  term of the A.P is zero   (HENCE PROVED) If you are satised do give thumbs up
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sorry i copied it credit goes to mhd imad 
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Thanks for this answer
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You are right it can never be zero
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In blue

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solution
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m+n th term is 0

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Seed and endiosperms
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If u know that m th term is term from last then solve it

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Member since Apr 11 2014 let the first be and common difference is . Given m times the mth term = n times the nth term Recall that the nth term of AP is Similarly, the mth term of AP is But, Hence, (m+n)th term of an AP is zero.
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mth term = a + (m-1)d
nth term = a + (n-1) d
Given that m{a +(m-1)d} = n{a + (n -1)d}
⇒ am + m²d -md = an + n²d - nd
⇒ am - an + m²d - n²d -md + nd = 0
⇒ a(m-n) + (m²-n²)d - (m-n)d = 0
⇒ a(m-n) + {(m-n)(m+n)}d -(m-n)d = 0
⇒ a(m-n) + {(m-n)(m+n) - (m-n)} d = 0
⇒ a(m-n)  + (m-n)(m+n -1) d  = 0
⇒ (m-n){a + (m+n-1)d} = 0
⇒ a + (m+n -1)d = 0/(m-n)
⇒ a + (m+n -1)d = 0
hence proved :)

 
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