Question

In an A.P of $n$ terms, the sum of first two terms is $b$ and the sum of the last two terms is $c$, then find the sum of its first $n$ terms.

Answer 1

We will assume the first term of the A.P as$a$ and the common difference as $d$. Now the A.P is $a,a+d,a+2d,a+3d,...$. In the problem they have mentioned the sum of the first two terms is $b$. So, we will calculate the sum of the first two terms of the A.P and equate it to the given value $b$, there we will get an equation. Again in the problem they have mentioned the sum of last two terms is $c$, so we will find the ${{n}^{th}}$ term and ${{\left( n-1 \right)}^{th}}$ term by using the known formula ${{a}_{n}}=a+\left( n-1 \right)d$ and calculate the sum of last two terms and equate it to $c$, here also we will get an equation. Now they have asked to calculate the sum of the $n$ terms, so the known formula for the sum of $n$ terms of A.P is ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. We will simplify the above equation by using the obtained equation in the above process to get the result.

Complete step by step answer:
Given that, an A.P has $n$ terms.
Let the first term of the A.P is $a$ and the common difference is $d$. Now the A.P is $a,a+d,a+2d,a+3d,...$.
The sum of the first two terms is $a+\left( a+d \right)=2a+d$.
In the problem they have mentioned that the sum of first two terms is $b$.
$\begin{aligned} & \therefore 2a+d=b \\\ & \Rightarrow 2a=b-d...\left( \text{i} \right) \\\ \end{aligned}$
We know that the ${{n}^{th}}$ term of the A.P is given by ${{a}_{n}}=a+\left( n-1 \right)d$ and the ${{\left( n-1 \right)}^{th}}$ term is $\begin{aligned} & {{a}_{n-1}}=a+\left( \left( n-1 \right)-1 \right)d \\\ & \Rightarrow {{a}_{n-1}}=a+\left( n-2 \right)d \\\ \end{aligned}$
Now the sum of the last two terms is given by
$\begin{aligned} & {{a}_{n}}+{{a}_{n-1}}=a+\left( n-1 \right)d+a+\left( n-2 \right)d \\\ & \Rightarrow {{a}_{n}}+{{a}_{n-1}}=2a+\left( n-1+n-2 \right)d \\\ & \Rightarrow {{a}_{n}}+{{a}_{n-1}}=2a+\left( 2n-3 \right)d \\\ \end{aligned}$
According to the given data the sum of the last two terms is $c$.
$\begin{aligned} & \therefore {{a}_{n}}+{{a}_{n-1}}=c \\\ & \Rightarrow 2a+\left( 2n-3 \right)d=c \\\ \end{aligned}$
From equation $\left( \text{i} \right)$ substituting the value $2a=b-d$ in the above equation, then we will get
$\begin{aligned} & b-d+\left( 2n-3 \right)d=c \\\ & \Rightarrow \left( 2n-3-1 \right)d=c-b \\\ & \Rightarrow \left( 2n-4 \right)d=c-b \\\ & \Rightarrow d=\dfrac{c-b}{2\left( n-2 \right)}...\left( \text{ii} \right) \\\ \end{aligned}$
We know that the sum of the $n$ terms in A.P is given by
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
From equation $\left( \text{i} \right)$ substituting the value $2a=b-d$ in the above equation, then we will get
$\begin{aligned} & {{S}_{n}}=\dfrac{n}{2}\left[ b-d+\left( n-1 \right)d \right] \\\ & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ b+\left( n-1-1 \right)d \right] \\\ & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ b+\left( n-2 \right)d \right] \\\ \end{aligned}$
From equation $\left( \text{ii} \right)$ substituting the value of $d=\dfrac{c-b}{2\left( n-2 \right)}$, then we will get
$\begin{aligned} & {{S}_{n}}=\dfrac{n}{2}\left[ b+\left( n-2 \right)\left( \dfrac{c-b}{2\left( n-2 \right)} \right) \right] \\\ & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ b+\dfrac{c-b}{2} \right] \\\ & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ \dfrac{2b+c-b}{2} \right] \\\ & \therefore {{S}_{n}}=\dfrac{n}{4}\left( b+c \right) \\\ \end{aligned}$

Note: While substituting the values of $2a$ and $d$. First substitute the value of $2a$ after that substitute the value of $d$, then we will get the result in an easy way. Otherwise If you substitute the values of $2a$ and $d$ at a time then you will need some time to simplify the equation.