Question

13. If m times the mth term of an A.P. is eqaul to n times nth term then show that the
(m + n) term of the A.P. is zero.​

Answer 1

Step-by-step explanation:

this is the answer okkkkkkkkkkkkkkkkkkkkk

Answer 2

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{ma_m \: = \: na_n} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  show :-  \begin{cases} &\sf{a_{(m + n)} = 0}  \end{cases}\end{gathered}\end{gathered}$

Cᴀʟᴄᴜʟᴀᴛɪᴏɴ :

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,

↝ mᵗʰ term is,

$\tt \:  \longrightarrow \: \begin{gathered}\bf{a_m\:=\:a\:+\:(m\:-\:1)\:d} \\ \end{gathered}$

Aɢᴀɪɴ,

↝ nᵗʰ term is,

$\tt \:  \longrightarrow \: \begin{gathered}\bf{a_n\:=\:a\:+\:(n\:-\:1)\:d} \\ \end{gathered}$

Nᴏᴡ, according to the statement,

$\tt \:  \longrightarrow \: m \: a_m \: = \: n \: a_n$

$\tt \:  \longrightarrow \: m \: (a + (m - 1)d) = n(a + (n - 1)d)$

$\tt \:  \longrightarrow \: m(a + md - d) = n(a + nd - d)$

$\tt \:  \longrightarrow \: am + d {m}^{2} - dm = an + {dn}^{2} - dn$

$\tt \:  \longrightarrow \: an + {dm}^{2} - dm - an - {dn}^{2} + dn = 0$

$\tt \:  \longrightarrow \: (am - an) + (d {m}^{2} - {dn}^{2} ) + (dn - dm) = 0$

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

$\tt \:  \longrightarrow \: a(m - n) + d(m - n)(m + n) - d(m - n) = 0$

$\tt \:  \longrightarrow \: (m - n)\bigg( a + (m + n)d - d\bigg) = 0$

$\tt\implies \:a + d(m + n) - d = 0 \: \: as \: m \: \ne \: n$

$\tt \:  \longrightarrow \: a \: + \: (m + n - 1)d = 0$

$\tt\implies \:a_{m \: + \: n} \: = \: 0$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$