Question

if m times the mth terms of AP is equals to n times its nth term then prove that its (m+n)th term is equal to zero

Answer 1

Ello user !!!!!!!!

Here is your answer,

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Let the first term of AP = a

common difference = d

We have to show that (m+n)th term is zero or a + (m+n-1)d = 0

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

Proved!

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Answer 2

Question : If m times the mth terms of AP is equal to n times its nth term, then prove that its (m+n)th term is equal to 0.

Answer :

Given :--

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

ma + m^2d - md = na + n^2d - nd

ma - na + m^2d - n^2d - md + nd = 0

a(m - n) +d(m^2 - n^2) - 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

Hence, proved.