Question

The first, second and nth term of the A.P. are $\alpha ,\beta \text{ and }\gamma $ respectively. Then the sum of the first n terms is
(a) $\beta +\gamma -2\alpha $
(b) $\dfrac{\beta +\gamma -2\alpha }{\beta -\alpha }$
(c) $\dfrac{\beta +\gamma +2\alpha }{\beta +\alpha }$
(d) $\dfrac{\left( \alpha +\gamma \right)\left( \beta +\gamma -2\alpha \right)}{2\left( \beta -\alpha \right)}$

Answer 1

Hint: In this question, we are given an Arithmetic progression and its first, second and nth term. Therefore, we can use the standard formula for the terms in an AP and get the common difference by subtracting the first term from the second term. Thereafter, we can use the formula for calculating the sum of n terms and express it in terms of $\alpha ,\beta \text{ and }\gamma $ to get the required answer to this question.

Complete step-by-step answer:
We know that the \[rth\] term of an Arithmetic progression is given by
${{a}_{r}}={{a}_{1}}+(r-1)d..............(1.1)$
Where ${{a}_{0}}$ is the first term and d is the common difference. As $\alpha ,\beta \text{ and }\gamma $ are given to be the first, second and \[nth\] term of the A.P., we can express them using equation (1.1) as
$\begin{align}
  & \alpha ={{a}_{1}}+(1-1)d \\
 & \Rightarrow {{a}_{1}}=\alpha ..............(1.2) \\
 & \beta ={{a}_{1}}+(2-1)d \\
 & \Rightarrow \beta ={{a}_{1}}+d..............(1.3) \\
\end{align}$
And
$\gamma ={{a}_{1}}+(n-1)d............(1.4)$
Using the value of ${{a}_{0}}$ from (1.2), in (1.3), we obtain
$\begin{align}
  & \beta =\alpha +d \\
 & \Rightarrow d=\beta -\alpha ...........(1.5) \\
\end{align}$
And using (1.5) and (1.2) in equation (1.4), we get
$\begin{align}
  & \gamma ={{a}_{1}}+(n-1)d=\alpha +(n-1)\left( \beta -\alpha \right) \\
 & \Rightarrow n-1=\dfrac{\gamma -\alpha }{\beta -\alpha } \\
 & \Rightarrow n=1+\dfrac{\gamma -\alpha }{\beta -\alpha }=\dfrac{\beta +\gamma -2\alpha }{\beta -\alpha }............(1.6) \\
\end{align}$
Now, the formula for finding the sum of n terms (${{s}_{n}}$ ) of the AP is given as
${{s}_{n}}=\dfrac{n}{2}\left( 2{{a}_{1}}+(n-1)d \right)=\dfrac{n}{2}\left( {{a}_{1}}+{{a}_{1}}+(n-1)d \right)$
Therefore, using the value of n from (1.6),${{a}_{0}}$ from (1.2), and $\gamma $ from in (1.4), we get
${{s}_{n}}=\dfrac{\beta +\gamma -2\alpha }{2\left( \beta -\alpha \right)}\left( \alpha +\gamma \right)$
Which matches option (d) in the given question. Hence, option (d) is the correct answer.

Note:We must note that the key point in the solution was to obtain the values of ${{a}_{0}}$, d and n from equations (1.2), (1.3) and (1.4). One can use any method to solve these equations to obtain the values of ${{a}_{0}}$, d and n and the answer will be same once we have the values of ${{a}_{0}}$, d and n regardless of the method used.