Question

if m times the mth term of an AP is equal to n times the nth term . show that (m+n)th term of the AP IS ZERO

Answer 1

The m th term of an A.P having first term a and common difference d is
$t_{m}$=a+(m-1)d and the n th term of the A.P is
$t_{n}$=a+(n-1)d
Then by the given condition,
m{a+(m-1)d}=n{a+(n-1)d}
or, am+m(m-1)d=an+n(n-1)d
or, a(m-n)+{m(m-1)-n(n-1)}d=0
or, a(m-n)+(m²-m-n²+n)d=0
or, a(m-n)+{(m+n)(m-n)-(m-n)}d=0
or, a(m-n)+{(m-n)(m+n-1)}d=0
or, (m-n){a+(m+n-1)d}=0
or, a+(m+n-1)d=0 [∵, m-n≠0 as m≠n]
or, a+{(m+n)-1}d=0
i.e., (m+n)th term of the A.P is 0. (proved)

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!