Question

if am=n and an=m then ap+m+n-p (prove) ​

Answer 1

Appropriate Question :-

In an AP, $\rm a_m = n \: and\: a_n = m,$ prove that $\rm a_p = m + n - p$

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

Let assume that

• First term of an AP = a

• Common difference of an AP = d

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ 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.

Tʜᴜs, According to statement,

$\rm \: a_m \: = \: n$

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

And

$\rm \: a_n = m$

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

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

$\rm \: (m - 1)d - (n - 1)d = n - m$

$\rm \: md -d - nd + d = n - m$

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

$\bf\implies \:d \: = \: - \: 1$

On substituting the value of d in equation (1), we get

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

$\rm \: a - m + 1 = n$

$\bf\implies \:a \: = \: n + m - 1$

Now, Consider

$\rm \: a_p$

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

On substituting the values of a and d, we get

$\rm \: = \: n + m - 1 + (p - 1)( - 1)$

$\rm \: = \: n + m - 1 - p + 1$

$\rm \: = \: n + m - p$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: a_p = \: n + m - p \: \: }}$

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

Additional Information :-

↝ 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.