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

Solution:

Here, we are given that $(m+1)^{\text {th }}$ term is twice the $(n+1)^{\text {th }}$ term, for a certain A.P. Here, let us take the first term of the A.P. as a and the common difference as d

We need to prove that $a_{3 m+1}=2 a_{m+n+1}$

So, let us first find the two terms.

As we know,

$a_{n^{\prime}}=a+\left(n^{\prime}-1\right) d$

For $(m+1)^{\text {th }}$ term $\left(n^{\prime}=m+1\right)$

$a_{m+1}=a+(m+1-1) d$

$=a+m d$

For $(n+1)^{\text {th }}$ term $\left(n^{\prime}=n+1\right)$

$a_{n+1}=a+(m+1-1) d$

$=a+m d$

For $(n+1)^{\text {th }}$ term $\left(n^{\prime}=n+1\right)$

$a_{n+1}=a+(n+1-1) d$

$=a+n d$

Now, we are given that $a_{m+1}=2 a_{n+1}$

So, we get,

$a+m d=2(a+n d)$

$a+m d=2 a+2 n d$

$m d-2 n d=2 a-a$

$(m-2 n) d=a$.......(1)

Further, we need to prove that the $(3 m+1)^{\text {th }}$ term is twice of $(m+n+1)^{\text {th }}$ term. So let us now find these two terms, For $(m+n+1)^{\text {th }}$ term $\left(n^{\prime}=m+n+1\right)$,

$a_{m+n+1}=a+(m+n+1-1) d$

$=(m-2 n) d+(m+n) d$

$=m d-2 n d+m d+n d$ (Using 1)

$=2 m d-n d$

For $(3 m+1)^{\text {th }}$ term $\left(n^{\prime}=3 m+1\right)$

$a_{3 m+1}=a+(3 m+1-1) d$

$=(m-2 n) d+3 m d$

$=m d-2 n d+3 m d$(Using 1)

$=4 m d-2 n d$

$=2(2 m d-n d)$

Therefore, $a_{3 m+1}=2 a_{m n+1}$

Hence proved