Question

If the sum of first m terms of an A.P. is same as the sum of its first (m+n) terms is zero.​

Answer 1

Appropriate Question :-

If the sum of first m terms of an A.P. is same as the sum of its first n terms, prove that sum of (m+n) terms is zero.

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

Let assume that

• First term of an AP = a

• Common difference of an AP = d

So, According to statement,

$\rm \: S_m = S_n$

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n  terms of an arithmetic progression 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 progression.

• n is the no. of terms.

• d is the common difference.

So, using this, we get

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

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

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

$\rm \: 2am + {dm}^{2} - dm = 2an + {dn}^{2} - dn$

$\rm \: 2am + {dm}^{2} - dm - 2an - {dn}^{2} + dn = 0$

$\rm \: 2a(m - n) + {d(m}^{2} - {n}^{2}) - d(m -n) = 0$

$\rm \: 2a(m - n) + d(m - n)(m + n) - d(m -n) = 0$

$\rm \:(m - n)\bigg[2a + d(m + n) - d\bigg] = 0$

As $\rm m \ne n$

$\bf\implies \:2a + (m + n - 1)d = 0 - - - (1)$

Now, Consider

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

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

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

$\rm \: = \: \dfrac{m + n}{2} \times 0$

$\rm \: = \: 0$

Hence,

$\bf\implies \: S_{m + n} \: = \: 0$

$\rule{190pt}{2pt}$

Additional Information :-

↝ nᵗʰ term of an arithmetic progression 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 progression.

• n is the no. of terms.

• d is the common difference.

Answer 2

$\huge\red{A}\pink{N}\orange{S}\green{W}\blue{E}\gray{R} =$

Appropriate Question :-

If the sum of first m terms of an A.P. is same as the sum of its first n terms, prove that sum of (m+n) terms is zero.

$\large\underline{\sf{Solution-}} </p><p>Solution−$

Let assume that

First term of an AP = a

Common difference of an AP = d

So, According to statement,

$\begin{gathered}\rm \: S_m = S_n \\ \end{gathered} </p><p>S </p><p>m</p><p></p><p> =S </p><p>n</p><p>$

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n terms of an arithmetic progression is,

$\begin{gathered}\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}\end{gathered} </p><p>★ </p><p>S </p><p>n</p><p></p><p> = </p><p>2</p><p>n</p><p></p><p> (2a+(n−1)d)$

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the progression.

n is the no. of terms.

d is the common difference.

So, using this, we get

$\begin{gathered}\rm \: \dfrac{m}{2} \bigg(2a + (m - 1)d \bigg) = \dfrac{n}{2} \bigg(2a + (n - 1)d\bigg) \\ \end{gathered} </p><p>2</p><p>m</p><p></p><p> (2a+(m−1)d)= </p><p>2</p><p>n</p><p></p><p> (2a+(n−1)d)$

$\begin{gathered}\rm \: m \bigg(2a + (m - 1)d \bigg) = n \bigg(2a + (n - 1)d\bigg) \\ \end{gathered} </p><p>m(2a+(m−1)d)=n(2a+(n−1)d)</p><p>$

$\begin{gathered}\rm \: 2am + m(m - 1)d = 2an + n(n - 1)d \\ \end{gathered} </p><p>2am+m(m−1)d=2an+n(n−1)d$

$\begin{gathered}\rm \: 2am + {dm}^{2} - dm = 2an + {dn}^{2} - dn \\ \end{gathered} </p><p>2am+dm </p><p>2</p><p> −dm=2an+dn </p><p>2</p><p> −dn$

$\begin{gathered}\rm \: 2am + {dm}^{2} - dm - 2an - {dn}^{2} + dn = 0 \\ \end{gathered} </p><p>2am+dm </p><p>2</p><p> −dm−2an−dn </p><p>2</p><p> +dn=0$

$\begin{gathered}\rm \: 2a(m - n) + {d(m}^{2} - {n}^{2}) - d(m -n) = 0 \\ \end{gathered} </p><p>2a(m−n)+d(m </p><p>2</p><p> −n </p><p>2</p><p> )−d(m−n)=0$

$\begin{gathered}\rm \: 2a(m - n) + d(m - n)(m + n) - d(m -n) = 0 \\ \end{gathered} </p><p>2a(m−n)+d(m−n)(m+n)−d(m−n)=0$

$\begin{gathered}\rm \:(m - n)\bigg[2a + d(m + n) - d\bigg] = 0 \\ \end{gathered} </p><p>(m−n)[2a+d(m+n)−d]=0</p><p>$

$As \rm m \ne nm</p><p></p><p>=n$

$\begin{gathered}\bf\implies \:2a + (m + n - 1)d = 0 - - - (1)\\ \end{gathered} </p><p>⟹2a+(m+n−1)d=0−−−(1)</p><p>$

Now, Consider

$\begin{gathered}\rm \: S_{m + n} \\ \end{gathered} </p><p>S </p><p>m+n$

$\begin{gathered}\rm \: = \: \dfrac{m + n}{2} \bigg(2a + (m + n - 1)d\bigg) \\ \end{gathered} </p><p>= </p><p>2</p><p>m+n</p><p></p><p> (2a+(m+n−1)d)$

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

$\begin{gathered}\rm \: = \: \dfrac{m + n}{2} \times 0 \\ \end{gathered} </p><p>= </p><p>2</p><p>m+n</p><p></p><p> ×0</p><p>$

$\begin{gathered}\rm \: = \: 0 \\ \end{gathered} </p><p>=0$

Hence,

$\begin{gathered}\bf\implies \: S_{m + n} \: = \: 0 \\ \end{gathered} </p><p>⟹S </p><p>m+n</p><p></p><p> =0</p><p></p><p> </p><p></p><p>\rule{190pt}{2pt}$

Additional Information :-

↝ nᵗʰ term of an arithmetic progression is,

$\begin{gathered}\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}\end{gathered} </p><p>★ </p><p>a </p><p>n</p><p></p><p> =a+(n−1)d</p><p></p><p> </p><p></p><p> </p><p></p><p> </p><p>,$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.
• a is the first term of the progression.
• n is the no. of terms.
• d is the common difference.