Question

How do you find the indicated sum for arithmetic series ${{S}_{15}}$ for $5+9+13+17+......?$

Answer 1

Hint: We will first recall the concept of Arithmetic Progression from sequence and series. We will use the formula ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$ to find the sum of first n terms of the sequence where n is the first term and d is the common difference of the two subsequent terms.

Complete step by step answer:
We know that arithmetic series also known as Arithmetic Progression is a type of the sequence where the difference between any two consecutive terms is constant. The general arithmetic Progression whose first term is a and common difference is d is given as: a, a+d, a+2d, a+3d, ……a+nd
So, for the arithmetic series given in the question, that is 5, 9, 13, 17, …:
First term will be equal to: 5.
And, the common difference is (9 - 5) = 4 = (13 - 9).
We can see from the question that we have to find ${{S}_{15}}$which is the sum of the first 15 terms of the series given:
And, for Arithmetic Progression, we know that the sum of the first n terms is given by:
${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$, where ‘a’ is the first term and ‘d’ is the common difference.
So, the sum of the first 15 terms of the series $5+9+13+17+......$ i.e. for n = 15 is equal to:
$\Rightarrow {{S}_{15}}=\dfrac{15}{2}\left( 2\times 5+\left( 15-1 \right)\times 4 \right)$
$\Rightarrow {{S}_{15}}=\dfrac{15}{2}\left( 10+56 \right)$
$\Rightarrow {{S}_{15}}=\dfrac{15}{2}\left( 66 \right)$
$\Rightarrow {{S}_{15}}=15\times 33$
$\Rightarrow {{S}_{15}}=495$

Hence, the sum of the first 15 terms of the series $5+9+13+17+......$ is equal to $495$.

Note: We can also find the sum alternatively by using the formula ${{S}_{n}}=\dfrac{n}{2}\left( a+l \right)$, where a is the first term of the series and ‘l’ is the last terms up to which we have to find the sum. So, we will first find the ${{15}^{th}}$ term of the series 5, 9, 13, 17, …: using the formula ${{T}_{n}}=a+\left( n-1 \right)d$, where a = 5 and n = 15 and d = 4.
So, ${{T}_{15}}=5+\left( 15-1 \right)4$
$\Rightarrow {{T}_{15}}=61$
Now, we will put the value of ${{1}^{st}}$ term, ${{15}^{th}}$ term and n = 15 in the formula ${{S}_{n}}=\dfrac{n}{2}\left( a+l \right)$ to get the required sum:
$\Rightarrow {{S}_{n}}=\dfrac{15}{2}\left( 5+61 \right)=15\times 33=495$
So, 495 is our required sum.