If (m+1)th term of an AP is twice the (n+1)th term,prove that (3m+1)th term is twice the (m+n+1)th term.

 Solution

Given that (m+1)th term of an AP is twice the (n+1)th term

Means, a(m+1)=2a(n+1) .............(i)

we know that  an= a+(n-1)d

use this in Equation (i)

a(m+1)=2a(n+1)

 a+(m+1-1)d = 2{a+(n+1-1)d}

a + md =2a+2nd

a -2a =+2nd-md

-a= d(2n-m)

a= -d(2n-m)

a=d(m- 2n)

we have to prove that (3m+1)th term is twice the (m+n+1)th term

means,  a(3m+1)=2a(m+n+1)

proof

LHS =

  a(3m+1)= a +(3m+1-1)d

a(3m+1)= a + 3md

but a=d(m- 2n)

so, a(3m+1)= d(m- 2n) + 3md

a(3m+1)= md - 2nd + 3md

a(3m+1)= 4md - 2nd (LHS)

again, RHS=2a(m+n+1)  

= 2{ a +(m+n+1-1)d}

 = 2{ a +(m+n)d}

 = 2a +2(m+n)d

 = 2a +2md+2nd

but a=d(m- 2n)  

= 2(d(m- 2n))+2md+2nd

= 2md- 4nd+2md+2nd

= 4md-2nd(RHS)

LHS = RHS