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Question: The first, second and nth term of the A.P. are \(\alpha ,\beta \text{ and }\gamma \) respectively. T...

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

Explanation

Solution

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 α,β and γ\alpha ,\beta \text{ and }\gamma to get the required answer to this question.

Complete step-by-step answer:
We know that the rthrth term of an Arithmetic progression is given by
ar=a1+(r−1)d..............(1.1){{a}_{r}}={{a}_{1}}+(r-1)d..............(1.1)
Where a0{{a}_{0}} is the first term and d is the common difference. As α,β and γ\alpha ,\beta \text{ and }\gamma are given to be the first, second and nthnth term of the A.P., we can express them using equation (1.1) as
α=a1+(1−1)d ⇒a1=α..............(1.2) β=a1+(2−1)d ⇒β=a1+d..............(1.3) \begin{aligned} & \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{aligned}
And
γ=a1+(n−1)d............(1.4)\gamma ={{a}_{1}}+(n-1)d............(1.4)
Using the value of a0{{a}_{0}} from (1.2), in (1.3), we obtain
β=α+d ⇒d=β−α...........(1.5) \begin{aligned} & \beta =\alpha +d \\\ & \Rightarrow d=\beta -\alpha ...........(1.5) \\\ \end{aligned}
And using (1.5) and (1.2) in equation (1.4), we get
γ=a1+(n−1)d=α+(n−1)(β−α) ⇒n−1=γ−αβ−α ⇒n=1+γ−αβ−α=β+γ−2αβ−α............(1.6) \begin{aligned} & \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{aligned}
Now, the formula for finding the sum of n terms (sn{{s}_{n}} ) of the AP is given as
sn=n2(2a1+(n−1)d)=n2(a1+a1+(n−1)d){{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),a0{{a}_{0}} from (1.2), and γ\gamma from in (1.4), we get
sn=β+γ−2α2(β−α)(α+γ){{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 a0{{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 a0{{a}_{0}}, d and n and the answer will be same once we have the values of a0{{a}_{0}}, d and n regardless of the method used.