Question

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

Answer 1

Step-by-step explanation:

$\begin{aligned} \rm t_{n} & \rm=a+(n-1) d \\ \\ \rm \text { Given } t_{m+1} & \rm=2 t_{n+1} \\ \\ \rm a+(m+1-1) d & \rm=2[a+(n+1-1) d] \\ \\ \rm a+m d & \rm=2(a+n d) \Rightarrow a+m d=2 a+2 n d \\ \\ \rm m d-2 n d & \rm=a \\ \\ \rm d(m-2 n) & \rm=a \quad \ldots .(1) \\ \\ \rm \text { To prove } t_{(3 m+1)} & \rm=2\left(t_{m+n+1}\right) \\ \\ \rm \text { L.H.S. } & \rm=t_{3 m+1} \\ \\ & \rm=a+(3 m+1-1) d \\ \\ & \rm=a+3 m d \\ \\ & \rm=d(m-2 n)+3 m d \quad \text { (from 1) } \\ \\ & \rm=m d-2 n d+3 m d \\ \\ & \rm=4 m d-2 n d \\ \\ & \rm=2 d(2 m-n) \end{aligned}$

$\[ \begin{aligned} \text { R.H.S. } & \rm=2\left(t_{m+n+1}\right) \\ \\ & \rm=2[a+(m+n+1-1) d] \\ \\ & \rm=2[a+(m+n) d] \\ \\ & \rm=2[d(m-2 n)+m d+n d)] \text { (from 1) } \\ \\ & \rm=2[d m-2 n d+m d+n d] \\ \\ & \rm=2[2 m d-n d] \\ \\ & \rm=2 d(2 m-n) \\ \\ \text { R.H.S } & =\text { L.H.S } \\ \\ \rm \therefore t_{(3 m+1)} & \rm=2 t_{(m+n+1)} \end{aligned} \]$

Hence it is proved.