Question

Solve for the data given below:
${{S}_{3}}=9,{{S}_{7}}=49$, find ${{S}_{n}}$ and ${{S}_{10}}$

Answer 1

Hint: We will apply the formula for finding the sum of first n terms of an arithmetic series. The formula for this is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ where, n are the terms which are selected for the sum. Also, a is the first term and d is the common difference between the terms of an Arithmetic series.

Complete step-by-step solution -
 Now we will first consider the sum of the first 3 terms of the arithmetic series. This is given to us as ${{S}_{3}}=9$.
By applying the formula ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ for sum of first n terms will result into ${{S}_{3}}=\dfrac{3}{2}\left[ 2a+\left( 3-1 \right)d \right]$ by keeping n = 3 where, n are the terms which are selected for the sum.
Also, a is the first term and d is a common difference. Therefore, after solving it we get ${{S}_{3}}=\dfrac{3}{2}\left[ 2a+2d \right]$.
Now, we will substitute the value of ${{S}_{3}}=9$ into the equation we get
$\begin{align}
  & 9=\dfrac{3}{2}\left[ 2a+2d \right] \\
 & \Rightarrow 9=3\left[ a+d \right] \\
 & \Rightarrow 9=3a+3d \\
 & \Rightarrow a+d=3…………………....(i) \\
\end{align}$
Now we will first consider the sum of first 7 terms of the arithmetic series. This is given to us as ${{S}_{7}}=49$.
By applying the formula ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ for sum of first n terms will result into ${{S}_{7}}=\dfrac{7}{2}\left[ 2a+\left( 7-1 \right)d \right]$ by keeping n = 9 where, n are the terms which are selected for the sum.
Also, a is the first term and d is a common difference. Therefore, after solving it we get ${{S}_{7}}=\dfrac{7}{2}\left[ 2a+6d \right]$.
Now, we will substitute the value of ${{S}_{7}}=49$ into the equation we get
$\begin{align}
  & 49=\dfrac{7}{2}\left[ 2a+6d \right] \\
 & \Rightarrow 7=\dfrac{1}{2}\left[ 2a+6d \right] \\
 & \Rightarrow a+3d=7………………………...(ii) \\
\end{align}$
Now we will solve equation (i) and equation (ii) by using elimination method. We will do this by subtracting the two equations as follows.
$\begin{align}
  & a+d=3 \\
 & \underline{\pm a\pm 3d=\pm 7} \\
 & \,\,\,\,\,-2d=-4 \\
\end{align}$
Therefore, we get $-2d=-4$. After simplifying this equation we get $d=2$. Now we will substitute this value of d in equation (i). Thus we get
$\begin{align}
  & a+d=3 \\
 & \Rightarrow a+2=3 \\
 & \Rightarrow a=3-2 \\
 & \Rightarrow a=1 \\
\end{align}$
Thus the values of a = 1 and d = 2.
Now, we will find the values of ${{S}_{n}}$ and ${{S}_{10}}$. For this we will first consider ${{S}_{n}}$ and use the formula ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. Therefore, after substituting a = 1 and d = 2 we get
$\begin{align}
  & {{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\
 & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2+\left( n-1 \right)2 \right] \\
 & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2+2n-2 \right] \\
 & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2n \right] \\
 & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\times 2n \\
 & \Rightarrow {{S}_{n}}={{n}^{2}} \\
\end{align}$
Now we will consider ${{S}_{10}}$ and use the formula ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. Therefore, after substituting a = 1 and d = 2 we get
$\begin{align}
  & {{S}_{10}}=\dfrac{10}{2}\left[ 2\left( 1 \right)+\left( 10-1 \right)2 \right] \\
 & \Rightarrow {{S}_{10}}=5\left[ 2+9\left( 2 \right) \right] \\
 & \Rightarrow {{S}_{10}}=5\left[ 2+18 \right] \\
 & \Rightarrow {{S}_{10}}=5\left[ 20 \right] \\
 & \Rightarrow {{S}_{10}}=100 \\
\end{align}$.

Note: We can also solve this directly by considering the sums ${{S}_{3}}=9,{{S}_{7}}=49$. Here, we can clearly see that ${{S}_{3}}=9$, which is actually the square of 3. Similarly, ${{S}_{7}}=49$ is the square of 7. By this we can directly have ${{S}_{n}}={{n}^{2}}$ and ${{S}_{10}}={{\left( 10 \right)}^{2}}$. While solving equations (i) and (ii) we can also use the method of substitution or Cramer’s rule to find the value of a and d. Instead of adding the equations (i) and (ii), we can also subtract them with each other. Focus while solving the equations otherwise, it may lead to wrong answers.