Question

If m times the mth term of an AP is equal to n times the nth term of same AP , prove that (m+n)th term of the AP is 0.

Answer 1

A.T.P,
m{ 2a +(m-1)d}=n{2a+(n-1)d}
m{2a+md-d}=n{2a+(n-1)d}
2am+m2d-md+2an+n2d-nd
2am-2an+m2d-n2d-md+nd=0   [m2 and n2 means "m" square and "n" square]
2a(m-n)+d(m2-n2)-d(m-n)+0
(m-n){2a+(m+n-1)d}=0
2a+(m+n-1)d=0
(m+n)th term=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!