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

`n^(th)` term in an AP is given by,
`T_n = a+(n-1)d`
`:. T_(m+1) = a+md`
`T_(n+1) = a+nd`
We are given,
`a+md = 2(a+nd)`
`=> a+md = 2a+2nd`
`=>a = (m-2n)d->(1)`
Now, `T_(3m+1) = a+3md`
From (1),
`T_(3m+1) = (m-2n)d+3md`
`= (m-2n+3m)d`
`=(4m-2n)d`
`T_(3m+1)=2(2m-n)d->(2)`
Now, `T_(m+n+1) = a+(m+n)d`
`=(m-2n)d+(m+n)d`
`=(m-2n+m+n)d`
`T_(m+n+1)=(2m-n)d->(3)`
From (2) and (3),
`T_(3m+1)= 2(T_(m+n+1))`