Question

11. If the first term of an A.P. is a, the second term
b and the last term c, then show that the sum of
the terms of the A.P. is
(a + c) (b + c -- 2a)/2 (b-a)​

Answer 1

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

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

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

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

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

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.

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n term of an arithmetic sequence is,

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

or

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

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 n terms.

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

Let d be the common difference of an AP.

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

$\tt : \implies \boxed{ \pink{ \bf \: \: d \: = \: b \: - \: a}}$

Now,

It is given that,

$\tt : \implies \: a_n \: = \: c$

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

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

On substituting the value of d, we get

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

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

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

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

$\tt : \implies \boxed{ \pink{ \bf \: n = \dfrac{c + b - 2a}{b - a} }}$

Now,

We know, that sum of n terms is given by

$\tt : \implies \: S_n \: = \dfrac{n}{2} (a_1 + a_n)$

$\rm \: On \: substituting \: the \: values \: of \: a_1, \: n, \: a_n, \: we \: get$

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

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