Question

How do you find the sum of the arithmetic series $1 + 3 + 5 + ....... + 27$?

Answer 1

Hint: In order to determine the sum, first find the total no of term in the A.P using nth term ${a_n} = a + (n - 1)d$ and later use ${S_n} = \dfrac{n}{2}\left\{ {2a + (n - 1)d} \right\}$ to calculate the required sum. Since, the last and first term of the sequence is given, we can alternatively use the formula ${S_n} = \dfrac{n}{2}\left\{ {a + l} \right\}$ to find out the value of sum.

Complete step by step answer:
Clearly, the given sequence is an Arithmetic Progression (A.P.) .
nth term of an A.P. is ${a_n} = a + (n - 1)d$
where a is the first term, d is the common constant difference
In our sequence first term $a = 1$ and difference $d = 3 - 1 = 2$.
So first we have to find the value of n i.e. total no of terms by putting ${a_n} = 27$
$
   \Rightarrow 27 = a + (n - 1)d \\
   \Rightarrow 27 = 1 + (n - 1)(2) \\
   \Rightarrow 27 = 1 + 2n - 2 \\
   \Rightarrow 27 = - 1 + 2n \\
   \Rightarrow 27 + 1 = 2n \\
   \Rightarrow 28 = 2n \\
   \Rightarrow n = \dfrac{{28}}{2} \\
   \Rightarrow n = 14 \\
 $
Total no of terms $n = 14$
Now to calculate the Sum of given A.P. series use the formula ${S_n} = \dfrac{n}{2}\left\{ {2a + (n - 1)d} \right\}$
 ${S_n} = \dfrac{n}{2}\left\{ {2a + (n - 1)d} \right\}$
Putting value of a, d and n
\[
  {S_n} = \dfrac{{14}}{2}\left\{ {2(1) + (14 - 1)(2)} \right\} \\
 \Rightarrow 7\left\{ {2 + 13(2)} \right\} \\
   \Rightarrow 7\left\{ {2 + 26} \right\} \\
   \Rightarrow 7\left\{ {28} \right\} \\
    \Rightarrow 196 \\
 \]
Therefore, the sum of the A.P. $1 + 3 + 5 + ....... + 27$ is equal to \[196\].

Additional Information:
1.Sequence: A sequence is a function whose domain is the set of N of natural numbers.
2.Real Sequence: A sequence whose range is a subset of R is called a real sequence.
In other words, a real sequence is a function having domain N and range equal to a subset of the set R of real numbers.
3.Arithmetic Progression (A.P): A sequence is called an arithmetic progression if the difference of a term and the previous term is always the same.
i.e. ${a_{n + 1}} - {a_n} = $ constant $( = d)$for all $n \in N$.
The constant difference is generally denoted by d which is called as the common difference.
In order to determine whether a sequence is an A.P. or not when its nth term is given, we may use the following algorithm .
Algorithm:
Step 1: Obtain ${a_n}$.
Step 2: Replace $n$ by $n + 1$in ${a_n}$ to get ${a_n} + 1$
Step 3: Calculate ${a_{n + 1}} - {a_n}$.
Step 4: If ${a_{n + 1}} - {a_n}$ is independent of n, the given sequence is an A.P. Otherwise is not an A.P. .

Note: You can alternatively use the formula of sum
${S_n} = \dfrac{n}{2}\left\{ {a + l} \right\}$
Where l is the last term, a is the first term, n is the total no of terms.
The difference between any two consecutive terms in an A.P. is always the same and if it is not the same, then the given series is not an A.P. $(n - 1)$ is nothing but the position of the term in the sequence.