Question

If m times the mth term of an AP is equal to n times its nth term, find (m+n)th term

Answer 1

Am = m [a + (m-1)d]

An = n[a + (n- 1)d]

Now (m + th term is given by

Am + An

[ma + m (m-1)d] + [na + n (n-1)d]

am + an + m(m-1)d + n (n-1)d

a (m+n) + [m (m-1) + n (n-1)]d


Thanks

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!