Question

If the sum of the first four terms of an AP is 40 and that of the first 14 terms is 280. Find the sum of its first n terms.

Answer 1

We start solving this problem by considering the formula for the sum of first $n$ terms of an AP with $a$ as its first term and $d$ as its common difference, that is, ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. Then we apply this formula for the first 4 terms and first 14 terms and we get two linear equations in terms of $a$ and $d$. We solve those two equations and get the values of $a$ and $d$. Finally, we substitute those values in the formula of the sum of first $n$ terms of an AP. Hence, we get the final result.

Complete step-by-step answer:
Let $a$ be the first term of the AP and $d$ be the common difference of the AP.
Now, let us consider the formula for the sum of first $n$ terms of an AP with $a$ as its first term and $d$ as its common difference, that is, ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$.
We were given that the sum of first four terms of the AP is 40.
By applying the above formula for 4 terms, we get,
$\begin{aligned} & {{S}_{4}}=\dfrac{4}{2}\left[ 2a+\left( 4-1 \right)d \right] \\\ & \Rightarrow \dfrac{4}{2}\left[ 2a+3d \right]=40 \\\ & \Rightarrow 2\left[ 2a+3d \right]=40 \\\ & \Rightarrow 2a+3d=\dfrac{40}{2} \\\ & \Rightarrow 2a+3d=20.....................\left( 1 \right) \\\ \end{aligned}$
We were also given that the sum of first 14 terms of the AP is 280.
By applying the above formula for 14 terms, we get,
$\begin{aligned} & {{S}_{14}}=\dfrac{14}{2}\left[ 2a+\left( 14-1 \right)d \right] \\\ & \Rightarrow \dfrac{14}{2}\left[ 2a+13d \right]=280 \\\ & \Rightarrow 7\left[ 2a+13d \right]=280 \\\ & \Rightarrow 2a+13d=\dfrac{280}{7} \\\ & \Rightarrow 2a+13d=40.....................\left( 2 \right) \\\ \end{aligned}$
Now, let us subtract the equation (1) from equation (2), we get,
$\begin{aligned} & \left( 2a+13d \right)-\left( 2a+3d \right)=40-20 \\\ & \Rightarrow 2a+13d-2a-3d=20 \\\ & \Rightarrow 13d-3d=20 \\\ & \Rightarrow 10d=20 \\\ & \Rightarrow d=\dfrac{20}{10} \\\ & \Rightarrow d=2 \\\ \end{aligned}$
Now, we substitute the value of $d$ in equation (1), we get,
$\begin{aligned} & 2a+3d=20 \\\ & \Rightarrow 2a+3\left( 2 \right)=20 \\\ & \Rightarrow 2a+6=20 \\\ & \Rightarrow 2a=20-6 \\\ & \Rightarrow 2a=14 \\\ & \Rightarrow a=\dfrac{14}{2} \\\ & \Rightarrow a=7 \\\ \end{aligned}$
Now, let us substitute the values of $a$ and $d$ in the formula of the sum of first $n$ terms of an AP.
$\begin{aligned} & {{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\\ & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2\left( 7 \right)+\left( n-1 \right)\left( 2 \right) \right] \\\ & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 14+2n-2 \right] \\\ & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2n+12 \right] \\\ & \Rightarrow {{S}_{n}}=n\left( \dfrac{2n+12}{2} \right) \\\ & \Rightarrow {{S}_{n}}=n\left( n+6 \right) \\\ \end{aligned}$
Therefore, the sum of first n terms of the given AP is $n\left( n+6 \right)$.
Hence, the answer is $n\left( n+6 \right)$.

Note: The possibilities of making mistakes in this type of problem is one may make a mistake by considering $a+\left( n-1 \right)d$ as the formula for sum of the first n terms instead of $\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. So, one must know the difference between the formula of ${{n}^{th}}$ term of the AP and sum of first n terms of the AP.