Question

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

Answer 1

Let a is the first term of AP and d is the common difference of AP
now ,
according to question ,
(m+1) th term =2 (n+1) th term
a+(m+1-1) d =2 {a+(n+1-1) d }
a+md=2a+2nd
(m-2n)d=a ------------(1)
now,
LHS =(3m+1) th term
=a+(3m+1-1) d
=(m-2n) d+3md
=2 (2m-n) d

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

hence LHS =RHS