Question

Consider the arithmetic sequence $3,7,11,........$
A. What is the sum of first n consecutive terms of the above sequence?
B. How many consecutive terms from the beginning should be added to get the sum 171?
C. Does the sum of the first few consecutive terms become 200? Why?

Answer 1

Hint: From the given series of arithmetic sequences, we find the common difference which is the difference between two consecutive terms. We put the values to get the formula for the general term ${{t}_{n}}$, the ${{n}^{th}}$ term of the series and the formula of summation. Then we put the value in the formula to find the solution.

Complete step-by-step solution:
We have been given a series of arithmetic sequence which is $3,7,11,........$
We express the arithmetic sequence in its general form.
We express the terms as ${{t}_{n}}$, the ${{n}^{th}}$ term of the series.
The first term be ${{t}_{1}}$ and the common difference be $d$ where $d={{t}_{2}}-{{t}_{1}}={{t}_{3}}-{{t}_{2}}={{t}_{4}}-{{t}_{3}}$.
We can express the general term ${{t}_{n}}$ based on the first term and the common difference.
The formula being ${{t}_{n}}={{t}_{1}}+\left( n-1 \right)d$.
The first term is 3. So, ${{t}_{1}}=3$. The common difference is $d={{t}_{2}}-{{t}_{1}}=7-3=4$.
We express general term ${{t}_{n}}$ as ${{t}_{n}}={{t}_{1}}+\left( n-1 \right)d=3+4\left( n-1 \right)=4n-1$.
Now we need to find the formula of summation of the arithmetic sequence.
The general formula for n terms is \[{{S}_{n}}=\dfrac{n}{2}\left[ 2{{t}_{1}}+\left( n-1 \right)d \right]\].
Therefore, sum of first n consecutive terms of the above sequence is
\[{{S}_{n}}=\dfrac{n}{2}\left[ 2\times 3+\left( n-1 \right)4 \right]=n\left( 2n+1 \right)=2{{n}^{2}}+n\].
For the sum to be 171, we get \[2{{n}^{2}}+n=171\]. We factorise and get
\[\begin{align}
  & 2{{n}^{2}}+n=171 \\
 & \Rightarrow 2{{n}^{2}}+n-171=0 \\
 & \Rightarrow 2{{n}^{2}}+19n-18n-171=0 \\
 & \Rightarrow \left( 2n+19 \right)\left( n-9 \right)=0 \\
\end{align}\]
For the value of $n$ we get $n=9$. Value of $n$ can’t be negative.
Therefore, we need 9 consecutive terms from the beginning to be added to get the sum 171.
Similarly, For the sum to be 200, we get \[2{{n}^{2}}+n=200\]. We factorise and get
\[\begin{align}
  & 2{{n}^{2}}+n=200 \\
 & \Rightarrow 2{{n}^{2}}+n-200=0 \\
 & \Rightarrow n=\dfrac{-1\pm \sqrt{1+1600}}{2}=\dfrac{-1\pm \sqrt{1601}}{2} \\
\end{align}\]
Value of $n$ can’t be irrational as it is an integer value.
Therefore, the sum can never be 200.

Note: The sequence is an increasing sequence where the common difference is a positive number. The common difference will never be calculated according to the difference of greater number from the lesser number.