Answer
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Hint: First find the common difference of the given arithmetic series by subtracting two successive terms. After finding the common difference “d” of the given arithmetic series use the below sum formula of arithmetic series in order to find the sum of the given arithmetic series: ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right],\;{\text{where}}\;{S_n}$ presents the sum of an arithmetic series up to $n$ terms having first term equals $a$ and common difference $d$
Complete step by step solution:
To find the sum of the arithmetic series $34 + 30 + 26 + .. + 2$ we will find the sum with help of formula for sum of $n$ numbers in arithmetic series, which is given as
${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Where ${S_n}$ is the sum of $n$ terms in an arithmetic series having $a$ as its first term and $d$ as a common difference.
Common difference $(d)$ can be determine by the subtraction of any two successive terms in series
$ \Rightarrow d = 26 - 30 = - 4$
So we get the common difference $d = - 4$
In the arithmetic series $34 + 30 + 26 + .. + 2$, we can see first term is $a = 34$
Also in an arithmetic series ${r^{{\text{th}}}}$ term is given as ${a_r} = a + (n - 1)d$
We will use this formula to find the value of $n$
We know that $2$ is the last term in $34 + 30 + 26 + .. + 2$
It can be written as
$
\Rightarrow 2 = 34 + (n - 1)( - 4) \\
\Rightarrow 2 = 34 - 4n + 4 \\
\Rightarrow 4n = 38 - 2 \\
\Rightarrow 4n = 36 \\
\Rightarrow n = 9 \\
$
Now putting all the values to get sum,
$
{S_9} = \dfrac{9}{2}\left[ {2 \times 34 + \left( {9 - 1} \right)( - 4)} \right] \\
= \dfrac{9}{2}\left[ {68 - 32} \right] \\
= \dfrac{{9 \times 36}}{2} \\
= 162 \\
$
$\therefore $ sum of the arithmetic series $34 + 30 + 26 + .. + 2\;{\text{is}}\;162$
Formula used:
1. Sum of Arithmetic series, ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
2. Formula for ${r^{{\text{th}}}}$ term, ${a_r} = a + (n - 1)d$
Note: Common difference tells about the nature of an arithmetic series, if common difference is positive then arithmetic series is increasing and if it is negative then arithmetic series is decreasing. The above formula is invalid for finding the sum of infinite arithmetic series.
Complete step by step solution:
To find the sum of the arithmetic series $34 + 30 + 26 + .. + 2$ we will find the sum with help of formula for sum of $n$ numbers in arithmetic series, which is given as
${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Where ${S_n}$ is the sum of $n$ terms in an arithmetic series having $a$ as its first term and $d$ as a common difference.
Common difference $(d)$ can be determine by the subtraction of any two successive terms in series
$ \Rightarrow d = 26 - 30 = - 4$
So we get the common difference $d = - 4$
In the arithmetic series $34 + 30 + 26 + .. + 2$, we can see first term is $a = 34$
Also in an arithmetic series ${r^{{\text{th}}}}$ term is given as ${a_r} = a + (n - 1)d$
We will use this formula to find the value of $n$
We know that $2$ is the last term in $34 + 30 + 26 + .. + 2$
It can be written as
$
\Rightarrow 2 = 34 + (n - 1)( - 4) \\
\Rightarrow 2 = 34 - 4n + 4 \\
\Rightarrow 4n = 38 - 2 \\
\Rightarrow 4n = 36 \\
\Rightarrow n = 9 \\
$
Now putting all the values to get sum,
$
{S_9} = \dfrac{9}{2}\left[ {2 \times 34 + \left( {9 - 1} \right)( - 4)} \right] \\
= \dfrac{9}{2}\left[ {68 - 32} \right] \\
= \dfrac{{9 \times 36}}{2} \\
= 162 \\
$
$\therefore $ sum of the arithmetic series $34 + 30 + 26 + .. + 2\;{\text{is}}\;162$
Formula used:
1. Sum of Arithmetic series, ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
2. Formula for ${r^{{\text{th}}}}$ term, ${a_r} = a + (n - 1)d$
Note: Common difference tells about the nature of an arithmetic series, if common difference is positive then arithmetic series is increasing and if it is negative then arithmetic series is decreasing. The above formula is invalid for finding the sum of infinite arithmetic series.
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