Question

If (m+1)th term of an Ap is twice the (n+1)th term. Prove that (3m+1)th term is twice the (n+n+1)th term.

Answer 1

let a and d be the first term and common diff for the A.P.

acc to qn

[a+[(m+1)-1]d=2[a+(n+1)-1]d

so

a+md=2a+2nd

so

a=(m-2n)d...(i)

now,

3m+1 th term

=a+[(3m+1)-1]d

=a+3md...

=4md-2nd (ii) using(I)

twice the m+n+1 th term

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

=2[a+(m+n)d]

=2[(m-2n)d+(m+n)d]

=4md-2nd..(iii)

from (ii) and (iii) the result is proved.