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

Explanation

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,
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