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.

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Vartika answered this

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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Mohd Imad answered this

Let mth term be a_m and nth term be a_n

According to the question :

m(a_m ) = n(a_n )

Apply Formula : a_n = 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] + { m² d- n²d } + (nd-md) = 0

a*[m-n] + d* { m² - n² } + d* (n-m) = 0

Apply : a² - 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 : a_n = a + (n-1) *d

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

So : a_m+n = 0

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

(HENCE PROVED)

If you are satisfied do give thumbs up.

Aman Wadhwa answered this

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

Vaibhav Mishra answered this

Given m × t_m = n × t_n
⇒ 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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Vaibhav Mishra answered this

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Vaibhav Mishra answered this

Dear Student, Thge general n^th 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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Vaibhav Mishra answered this

) 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

Vaibhav Mishra answered this

mt_m = nt_n(Given) ∴ m [a + (m – 1) d] = n [a + (n – 1) d] [∵ t_n = 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) t_m+n = a + (m + n – 1) [∵ t_n = a + (n – 1) d] ∴ t_{m + n} = 0 [from eq. no. (1)] Hence proved.

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Hemang Rattra answered this

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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Jyoti Ranjan answered this

if a(m+n)=0 then prove that (m+n)th term is 0

Megharaj answered this

Jahnavi Karumanthra answered this

thank you so much

Aaditya answered this

thumbs up

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Ashish answered this

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

Ashish answered this

sorry i copied it credit goes to mhd imad

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Lovisha answered this

Thanks for this answer

Parteek answered this

You are right it can never be zero

Sambhav Goyal answered this

In blue

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Priya Singh answered this

solution

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Priya Singh answered this

m+n th term is 0

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Seed and endiosperms

Diwakar answered this

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

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Vasudevan Ks answered this

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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Aayush.k answered this

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 :)