Question

if m times the mth term of an ap is equal to n times the nth term and n is not equal to n .show that its (m+n)th termis zero

Answer 1

m*(a+(m-1)d)=n*(a+(n-1)d)
am+md(m-1)=an+nd(n-1)
am+$m^{2}$d-md=an+$n^{2}$d-nd

now bringing everything to LHS,
am-an+m^2d-md-n^2d+nd=0
a(m-n)+d(m^2-n^2-m+n)=0
a(m-n)+d((m+n)*(m-n)-(m-n))=0
(m-n)(a+(m+n-1)d=0
=>a+(m+n-1)=0
amn=0

Answer 2

Answer:

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!