Math, asked by NITESH761, 1 month ago

if  \rm (m+1)^{th} term of an AP is twice the  \rm (n+1)^{th} term, prove that
 \rm (3m+1)^{th} term is twice the  \rm (m+n+1)^{th} term.​

Answers

Answered by mathdude500
3

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

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}

Wʜᴇʀᴇ,

  • aₙ is the nᵗʰ term.

  • a is the first term of the sequence.

  • n is the no. of terms.

  • d is the common difference.

Tʜᴜs,

According to statement

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

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

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

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

\bf\implies \:a = md - 2nd -  -  - (1)

Now, We have to prove that

\red{\rm :\longmapsto\:\boxed{ \tt{ \: a_{3m + 1} = 2a_{m + n + 1}} \: }}

So, Consider

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

\rm \:  =  \: a + (3m + 1 - 1)d

\rm \:  =  \: a + 3md

On substituting the value of a from equation (1), we get

\rm \:  =  \: md - 2nd + 3md

\rm \:  =  \: 4md - 2nd

\rm \implies\:\boxed{ \tt{ \: a_{3m + 1} \: =  \: 2(2md - nd) \: }} -  -  - (2)

Now, Consider

\rm :\longmapsto\:a_{m + n + 1}

\rm \:  =  \: a + (m + n + 1 - 1)d

\rm \:  =  \: a + (m + n)d

On substituting the value of a from equation (1), we get

\rm \:  =  \: md - 2nd + dm + nd

\rm \:  =  \: 2md - nd

\rm \implies\:\boxed{ \tt{ \: a_{m +n +  1} \: =  \: 2md - nd}} -  -  - (3)

Hence, from equation (2) and (3), we concluded that

\red{\rm :\longmapsto\:\boxed{ \tt{ \: a_{3m + 1} = 2a_{m + n + 1}} \: }}

Hence, Proved

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More to know :-

↝ Sum of n  terms of an arithmetic sequence is,

\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}

Wʜᴇʀᴇ,

  • Sₙ is the sum of n terms of AP.

  • a is the first term of the sequence.

  • n is the no. of terms.

  • d is the common difference.
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