Question

If ${{S}_{n}}$ denotes the sum of the first $n$ terms of an A.P prove that ${{S}_{12}}=3\left( {{S}_{8}}-{{S}_{4}} \right)$ .

Answer 1

For these kinds of questions, we need to know the concepts of Arithmetic Progressions or in short known as AP. Arithmetic Progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms in constant. The first term in the series or the sequence is usually denoted by the letter $a$ and the common difference between the consecutive numbers is denoted by $d$. The following terms after the first term are denoted by ${{a}_{n}}$ where $n=2,3,4.....n$.

Complete step by step solution:
The terms after the first term is denoted by ${{a}_{n}}$ where $n=2,3,4.....n$. As we plug in the value of $n$, we get each term. ${{a}_{1}}$ is same as $a$.
So ${{a}_{2}}$ is the second term, ${{a}_{3}}$ is the third term and so on.
The last term of any sequence which has $n$ terms is denoted by ${{a}_{n}}$ .
If a sequence has $m$ terms, then the last term would be ${{a}_{m}}$ .
There is already a derived formula to find out the last term of an AP. It is the following :
$\Rightarrow {{a}_{n}}=a+\left( n-1 \right)d$
Let us see the formula to find the sum of the $n$ terms of an AP .It is denoted by ${{S}_{n}}$ . The formula is the following :
$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$
Let us distribute the $a$ in the formula.
Upon doing so, we get the following :
$\begin{aligned} & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right) \\\ & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( a+a+\left( n-1 \right)d \right) \\\ \end{aligned}$
It is nothing but the formula to find the last term of an AP.
Let us shrink it. Upon doing so, we get the following :

& \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right) \\\ & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( a+a+\left( n-1 \right)d \right) \\\ & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( a+{{a}_{n}} \right) \\\ \end{aligned}$$ Let us replace ${{a}_{n}}$ with $l$ so that it represents the last term since ${{a}_{n}}$ is the last term. $$\begin{aligned} & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right) \\\ & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( a+a+\left( n-1 \right)d \right) \\\ & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( a+{{a}_{n}} \right) \\\ & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( a+l \right) \\\ \end{aligned}$$ Now let us calculate ${{S}_{12}}$. It is nothing but the sum of the first $12$ terms of an AP. Let us substitute $n=12$. Upon doing so, we get the following : $$\begin{aligned} & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( a+l \right) \\\ & \Rightarrow {{S}_{12}}=\dfrac{12}{2}\left( a+l \right) \\\ & \Rightarrow {{S}_{12}}=6\left( a+l \right) \\\ \end{aligned}$$ Now let us calculate ${{S}_{8}}$. It is nothing but the sum of the first $8$ terms of an AP of the same AP. Since it is the same AP, the last term $l$ would be the same. Let us substitute $n=8$. Upon doing so, we get the following : $$\begin{aligned} & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( a+l \right) \\\ & \Rightarrow {{S}_{8}}=\dfrac{8}{2}\left( a+l \right) \\\ & \Rightarrow {{S}_{8}}=4\left( a+l \right) \\\ \end{aligned}$$ Now let us calculate ${{S}_{4}}$. It is nothing but the sum of the first $4$ terms of an AP of the same AP. Since it is the same AP, the last term $l$ would be the same. Let us substitute $n=4$. Upon doing so, we get the following : $$\begin{aligned} & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( a+l \right) \\\ & \Rightarrow {{S}_{4}}=\dfrac{4}{2}\left( a+l \right) \\\ & \Rightarrow {{S}_{4}}=2\left( a+l \right) \\\ \end{aligned}$$ Now let us substitute and solve $3\left( {{S}_{8}}-{{S}_{4}} \right)$. Upon doing so, we get the following : $\begin{aligned} & \Rightarrow 3\left( {{S}_{8}}-{{S}_{4}} \right)=3\left( 4\left( a+l \right)-2\left( a+l \right) \right) \\\ & \Rightarrow 3\left( {{S}_{8}}-{{S}_{4}} \right)=3\left( 4a+4l-2a-2l \right) \\\ & \Rightarrow 3\left( {{S}_{8}}-{{S}_{4}} \right)=3\left( 2a+2l \right) \\\ & \Rightarrow 3\left( {{S}_{8}}-{{S}_{4}} \right)=6\left( a+l \right) \\\ & \Rightarrow 3\left( {{S}_{8}}-{{S}_{4}} \right)={{S}_{12}} \\\ \end{aligned}$ $\therefore $ Hence proved. **Note:** It is very important to know all the definitions and formulae of Arithmetic Progressions. It is important to remember all the short cuts too. We should be careful while solving as there is a lot of scope for calculation errors. We should have a lot of practice in this chapter. Arithmetic Mean is quite important in this chapter. We should also learn all the formulae relating Geometric Progression and Harmonic Progression. We should also know the relationship between all these three.