Question

$\quad \leadsto \sf AP = a_{1} , a_{2} , . . . . . . , a_{n}$

$\quad \leadsto \sf S_{n} = m$

$\quad \leadsto \sf S_{m} = n$

Fínd $\sf S_{m+n}$

May be a good question -,- ​

Answer 1

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

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ 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.

Given that,

$\red{\rm :\longmapsto\:S_n = m}$

$\rm :\longmapsto\:\dfrac{n}{2}\bigg(2a + (n - 1)d\bigg) = m$

$\rm :\longmapsto\:2an + n(n - 1)d = 2m - - - (1)$

Further, given that

$\red{\rm :\longmapsto\:S_m = n}$

$\rm :\longmapsto\:\dfrac{m}{2}\bigg(2a + (m - 1)d\bigg) = n$

$\rm :\longmapsto\:2am + m(m - 1)d =2n - - - (2)$

On Subtracting equation (1) from (2), we get

$\rm :\longmapsto\:2a(m - n) + d\bigg(m(m - 1) - n(n - 1)\bigg) = 2n - 2m$

$\rm :\longmapsto\:2a(m - n) + d\bigg( {m}^{2} - m - {n}^{2} + n\bigg) = 2n - 2m$

$\rm :\longmapsto\:2a(m - n) + d\bigg({m}^{2} -{n}^{2} + n - m\bigg) = 2n - 2m$

$\rm :\longmapsto\:2a(m - n) + d\bigg((m + n)(m - n) - (m - n)\bigg) = 2n - 2m$

$\rm :\longmapsto\:2a(m - n) + d(m - n)\bigg(m + n - 1\bigg) = 2n - 2m$

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

$\rm :\longmapsto\:2a+ d\bigg(m + n - 1\bigg)= - 2 - - - - (3)$

Now, Consider

$\rm :\longmapsto\:S_{m + n}$

$\rm \:  =  \: \dfrac{m + n}{2}\bigg(2a + (m + n - 1)d\bigg)$

$\rm \:  =  \: \dfrac{m + n}{2}\bigg( - 2\bigg)$

$\rm \:  =  \: - (m + n)$

Hence,

$\rm \implies\:\boxed{\tt{ \:S_{m + n} \: = \: - \: (m + n) \: }}$

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Additional Information

↝ 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.

Answer 2

given★

• $\quad \leadsto \sf AP = a_{1} , a_{2} , . . . . . . , a_{n}$

• $\quad \leadsto \sf S_{n} = m$

• $\quad \leadsto \sf S_{m} = n$

find★

• $\sf S_{m+n}$

solution★

• we know that sum of n th term formula

• $s_{n} = \frac{n}{2} (2a + (n - 1)d)$

• $\quad \leadsto \sf S_{n} = m$

• $\frac{n}{2} (2a + (n - 1)d) = m \\ \\ \\ 2an + nd(n - 1)d = 2m - - - (1)$

• $\quad \leadsto \sf S_{m} = n$

• similarly

• $\frac{m}{2} (2a +( m - 1)d = n$

• $2am + md(m - 1) = 2n - - - (2)$

• now we subsitude (1)-(2)

• $2an - 2am +nd(n - 1) - md(m - 1) = 2n - 2m$

• $2a(n - m) + d(n(n - 1) - m(m - 1)) = 2(n - m)$

• $(n - m)(2a ( - m - n + 1)d) = 2(n - m) \\ \\ 2a - (- m - n + 1)d = - 2 \\ \\ 2a + (m + n - 1)d = - 2 - - - (3)$

============================≠

• $s_{n + m} = \frac{n + m}{2}(2a + (m + n - 1)d) \\ \\ \\ eq \: \: (3) value \: put \\ \\ \\ = \frac{m + n}{2} ( - 2) \\ \\ \\ - (m + n)$

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other formula

• $s_{n} = \frac{n}{2} (2a + l)$
• where l last term number

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I hope it helps you