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.

with step by step explanation​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

• First term of an AP = a

• Common difference of an AP = d

Now, it is given that (m+1)th term of an A.P. is twice the (n+1)th term.

$\rm \: a_{m + 1} \: = \: 2a_{n + 1}$

$\rm \: a + (m + 1- 1)d = 2[a + (n + 1 - 1)d]$

$\rm \: a + md = 2a + 2nd$

$\rm \: a = md - 2nd$

$\rm\implies \:\rm \: a = (m - 2n)d$

Now, we have to prove that, (3m+1)th term is twice the (m+n+1)th term.

i.e.

$\bf \: a_{3m + 1} \: = \: 2a_{m + n + 1}$

Now, Consider

$\rm \: a_{3m + 1}$

$\rm \: = \: a + (3m + 1 - 1)d$

$\rm \: = \: a + 3md$

On substituting the value of a, we have

$\rm \: = \: (m - 2n)d+ 3md$

$\rm \: = \: (m - 2n+ 3m)d$

$\rm \: = \: (4m - 2n)d$

$\bf\implies \:a_{3m + 1} = \: (4m - 2n)d - - - (1)$

Now, Consider

$\rm \: 2a_{m + n + 1}$

$\rm \: = \: 2[a + (m + n + 1 - 1)d]$

On substituting the value of a, we get

$\rm \: = \: 2[(m - 2n)d + (m + n)d]$

$\rm \: = \: 2[(m - 2n + m + n)d]$

$\rm \: = \: 2[(2m - n )d]$

$\rm \: = \: (4m - 2n)d$

$\bf\implies \:2a_{m + n + 1} = \: (4m - 2n)d - - - (2)$

From equation (1) and (2), we concluded that

$\rm\implies \:\boxed{ \rm{ \:\bf \: a_{3m + 1} \: = \: 2a_{m + n + 1} \: }}$

Answer 2

$\small\colorbox{lightyellow} {\text{ \bf♕ Brainliest answer }}$

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$\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}$

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

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