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

m(mth term)=n(nth term)
m(a+(m-1)d)=n(a+(n-1)d)
am+md(m-1)=na+nd(n-1)
a(m-n)=d(n²-n-m²+m)
a(m-n)=d(m-n-(m²-n²))
a(m-n)=d(m-n)(1-m-n)
a=d(1-m-n)

(m+n)th term=a+(m+n-1)d=d(1-m-n)+(m+n-1)d
=d(1-m-n+m+n-1)
=d*0=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!