Question

Find the sum of the first 20 terms of the arithmetic series in which ${{3}^{rd}}$ term is 7 and ${{7}^{rd}}$ term is 2 more than three times its ${{3}^{rd}}$ term.

Answer 1

Hint: We first try to find the general equations and series for an A.P. We assume the first term and the common difference. We find the equation of general term, ${{3}^{rd}}$ term is 7 and ${{7}^{rd}}$ term is 2+3$T_3$. Then we put the values of 3 and 7 in place of the equations and find the relation.

Complete step-by-step solution:
We express the A.P. in its general form.
We express the terms as ${{t}_{n}}$, the ${{n}^{th}}$ term of the series.
Let the first term be a. So, ${{t}_{1}}=a$. The common difference is d. So, $d={{t}_{2}}-{{t}_{1}}={{t}_{n}}-{{t}_{n-1}}$.
We express the general term ${{t}_{n}}$ as ${{t}_{n}}=a+\left( n-1 \right)d$.
The sum of n terms will be ${{S}_{n}}$ as ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. We need to find ${{S}_{20}}$.
It’s given that ${{3}^{rd}}$ term is 7 and ${{7}^{rd}}$ term is 2 more than three times its ${{3}^{rd}}$ term which means ${{7}^{rd}}$ term is \[3\times 7+2=23\].
So, ${{t}_{3}}=a+\left( 3-1 \right)d\Rightarrow a+2d=7....(i)$ and ${{t}_{7}}=a+\left( 7-1 \right)d\Rightarrow a+6d=23....(ii)$
We got 2 equations of two variables and we solve them
We subtract equation (i) from equation (ii) to get
$\begin{align}
  & \left( a+6d \right)-\left( a+2d \right)=23-7 \\
 & \Rightarrow 4d=16 \\
 & \Rightarrow d=\dfrac{16}{4}=4 \\
\end{align}$
Putting the value of d in equation (i) we get value of a.
$\begin{align}
  & \Rightarrow a+2\times 4=7 \\
 & \Rightarrow a=7-8=-1 \\
\end{align}$
So, the series is -1, 3, 7, 11, ….
The sum of 20 terms will be ${{S}_{n}}=\dfrac{20}{2}\left[ 2\left( -1 \right)+\left( 20-1 \right)4 \right]=10\times 74=740$.
Therefore, the sum of the first 20 terms of the arithmetic series is 740.

Note: The series is actually an ascending series with difference 4. Although the series starts with a negative number, the only negative number is the first term. The sum ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ can also be expressed as the ${{S}_{n}}=\dfrac{n}{2}\left[ a+a+\left( n-1 \right)d \right]=\dfrac{n}{2}\left[ {{t}_{1}}+{{t}_{n}} \right]$.