Question

if [n+1]th term of a.p s twice the [m+1]th term. prove that [3m+1]th term is twice that [m+n+1]th term?

Answer 1

Am+1=28An+1
 to proof.A3m+1=2Am+n+1
     a+(m+1-1)=2[a+(n+1-1)d]
a+md=2(a+nd)
a+md=2a+2nd
2a-a+2nd-md=0
a+(2n-m)d=0
a=-(2n-m)d=(m-2n)d
L.H.S.=A3m=1
=a+3m+1-1)d
=(m-2n)d+3md
=(m-2n+3m)d
=(4m-2n)d

R.H.S.
 2Am+n+1=2[a+(m+n+1-1)d]
2[a(m-2n)d+(m+n)d
=2(m-2n+m+n)d
2(2m-n)d
=(4m-2n)d
hence proved LHS=RHS

Answer 2

hey.....try to see this uploaded image.....i have tried to solve it in less steps....