Question

M times the mth term is equal to n times the n th term. Then prove that m+n=0.

Answer 1

question ask
mtm=ntn
where tm and tn are mth and nth terms of AP
let this AP first term is a and common difference d
then
m {a+(m-1) d}=n {a+(n-1) d}
a (m-n)+d (m^2-n^2)-d (m-n)=0
(m-n){a+(m+n-1) d}=0

{a+(m+n-1)d}=0

t (m+n)=0. hence proved

Answer 2

Answer:

Step-by-step explanation: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!