Question

If n Times the nth term an ap is equal to m times it's mth term, the (m+n)th term is

Answer 1

Answer:

Let first term a and common difference d

$n(a + (n - 1)d) = m(a + (m - 1)d) \\ (m - n)a + (m - n)d = 0 \\ (m - n)(a + d) = 0 \\ \\ m + n \: th \: \: term \\ = a + (m + n - 1)d \\ = - d + (m + n - 1)d \\ = (m + n - 2)d$

We stuck here . May be you meant m times nth term . Recheck question please

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!