Question

The‌ ‌first,‌ ‌second‌ ‌and‌ ‌middle‌ ‌terms‌ ‌of‌ ‌an‌ ‌AP‌ ‌are‌ ‌a,‌ ‌b,‌ ‌c‌ ‌respectively.‌ ‌Their‌ ‌sum‌ ‌is‌ ‌

Answer 1

Hint: Before solving the above let's discuss the AP or arithmetic progression. In mathematics, an arithmetic progression or AP or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

Complete step by step solution:
If the initial term of the arithmetic progression is ${{a}_{1}}$and the common difference of successive members is $d$, then the $nth$ term of the sequence $\left( {{a}_{n}} \right)$ is given by:
$\Rightarrow {{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$ and we can also write if we consider $m=1$ then in general way we can write
$\Rightarrow {{a}_{n}}={{a}_{m}}+\left( n-m \right)d$
A finite portion of an arithmetic progression is called a finite arithmetic progression and the sum of a finite arithmetic progression is called an arithmetic series. Now we know that the general formula to find the sum of the arithmetic progression is
$\Rightarrow S=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$
The middle term in AP is exist when $n\to odd$ otherwise the middle term is become zero
Now we know,
$\Rightarrow d=b-a$
Here we can also write
$\Rightarrow c=a+\left( \dfrac{n+1}{2}-1 \right)d$
Or we can write
$\begin{align}
  & \Rightarrow c=a+\left( \dfrac{n-1}{2} \right)\left( b-a \right) \\
 & \Rightarrow 2c=2a+\left( n-1 \right)\left( b-a \right) \\
\end{align}$
Now find the value of$n$from the above equation, we get
$\begin{align}
  & \Rightarrow n-1=\dfrac{2c-2a}{b-a} \\
 & \Rightarrow n=\dfrac{2c+b-3a}{b-a} \\
\end{align}$
Now we know that the sum of the terms $S=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$, put the value of $n$ in this sum formula we get,
$\begin{align}
  & \Rightarrow S=\dfrac{2c+b-3a}{2\left( b-a \right)}\left[ 2a+\left( \dfrac{2c-2a}{b-a} \right)\left( b-a \right) \right] \\
 & \Rightarrow S=\dfrac{2c+b-3a}{2\left( b-a \right)}\left( 2c \right) \\
\end{align}$
More simplifying it we get,
$\begin{align}
  & \Rightarrow S=\left( \dfrac{2c+b-3a}{b-a} \right)c \\
 & \Rightarrow S=\dfrac{2c\left( c-a \right)}{b-a}+c \\
\end{align}$
Hence we get the sum of the arithmetic progression is $S=\dfrac{2c\left( c-a \right)}{b-a}+c$.

Note: To solve these types of the question we should know everything about the arithmetic progression. If we observe in our regular lives, we come across arithmetic progression quite often.