Question

sʜᴏᴡ ᴛʜᴀᴛ ᴛʜᴇ sᴜᴍ ᴏғ ᴀʟʟ ᴛᴇʀᴍs ᴏғ ᴀɴ ᴀ.ᴘ. ᴡʜᴏsᴇ ғɪʀsᴛ ᴛᴇʀᴍ ɪs a ,ᴛʜᴇ sᴇᴄᴏɴᴅ ᴛᴇʀᴍ ɪs b ᴀɴᴅ ᴛʜᴇ ʟᴇᴛ ᴛᴇʀᴍ ɪs c ɪs ᴇǫᴜᴀʟ ᴛᴏ
$\frac{(a + c)(b + c - 2a)}{2(b - a)}$
​

Answer 1

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{a_1 = a} \\ &\sf{a_2 = b}\\ &\sf{a_n = c} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  prove :-  \begin{cases} &\sf{S_n = \dfrac{(a + c)(b + c - 2a)}{2(b - a)} }  \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

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

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

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} (\:a\:+ \: a_n)}}}}}} \\ \end{gathered}$

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.

• Sₙ is the sum of first n terms

$\large\underline\purple{\bold{Solution :- }}$

Given That :-

$\tt \:  \longrightarrow \: a_1 = a$

$\tt \:  \longrightarrow \: a_2 = b$

$\tt\implies \:d \: = a_2 \: - \: a_1$

$\tt\implies \: \boxed{ \red{\tt \:  d \: = \: b \: - \: a \: - - (i)}}$

$\begin{gathered}\bf\pink{Now} \\ \end{gathered}$

$\tt \:  \longrightarrow \: a_n \: = \: c$

$\tt\implies \:a \: + \: (n \: - 1)d \: = \: c$

$\tt \:  \longrightarrow \: (n - 1)(b - a) = c - a \: \: \: (using \: (i))$

$\tt\implies \:n \: - \: 1 \: = \dfrac{c - a}{b - a}$

$\tt\implies \:n \: = \: \dfrac{c - a}{b - a} + 1$

$\tt\implies \:n \: = \: \dfrac{c - a + b - a}{b - a}$

$\tt\implies \:n \: = \: \dfrac{c + b \: - \: 2 a}{b - a} - - (ii)$

$\begin{gathered}\bf\pink{Now} \\ \end{gathered}$

$\tt \:  \longrightarrow \: S_n \: = \dfrac{n}{2} (a \: + \: a_n)$

☆ On substituting the values, we get

$\tt \:  \longrightarrow \: S_n = \dfrac{(b \: + \: c \: - \: 2a)}{2(b \: - \: a)} (a \: + \: c)$

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