Question

Find the sum of the following sequential series $0 - 1 + 2 - 3 + 4 - 5 + 6 - 7........ + 16 - 17 + 18 - 19 + 20$ respectively.
(a) $80$
(b) $10$
(c) $30$
(d) $50$

Answer 1

Hint: The given problem revolves around the concepts of Arithmetic Progression (A.P.). First of all separating the negative terms and positive terms, two A.P. series are formed. As a result, finding the total number of factors ‘n’ for both the series by using the ‘${t_n} = a + (n - 1)d$’ formula. Then, using the formula to calculate the sum i.e. ${S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]$ of the certain kind sequence, the desired value or sum is obtained.

Complete step-by-step solution:
Since, we have given the sequence that
$0 - 1 + 2 - 3 + 4 - 5 + 6 - 7........ + 16 - 17 + 18 - 19 + 20$
As a result, considering the sum of sequence as ‘Total sum’, we get
$Total{\text{ }}sum = 0 - 1 + 2 - 3 + 4 - 5 + 6 - 7........ + 16 - 17 + 18 - 19 + 20$
As per the given sequence is concerned, mathematically the series shows the Arithmetic Progression (A.P.) having definite common difference with first term as well as last term respectively.
Solving the sequence mathematically that is taking the negative terms in one bracket, we get
$Total{\text{ }}sum = - \left( {1 + 3 + 5 + ..... + 17 + 19} \right) + \left( {2 + 4 + 6 + ..... + 18 + 20} \right)$ …………………………… (i)
Now, we will find the sum individually
Hence,
Considering the first sequence i.e. $\left( {1 + 3 + 5 + ..... + 17 + 19} \right)$
It seems that, $[{t_n} = a + (n - 1)d]$ represents total terms in the respective sequence
Where,
‘${t_n}$’ is the last term of the arithmetic series, i.e. ${t_n} = 19$
‘a’ is the first term of the sequence or series, i.e. $a = 1$
‘n’ is the terms in the respective series or sequence,
‘d’ is the common difference of the arithmetic series expressed as $[d = {t_2} - {t_1}]$ where, ${t_2}$is second term and \[{t_1}\] is first term of the series respectively, i.e. $d = {t_2} - {t_1} = 3 - 1 = 2$
Now, substituting the values in the equation $[{t_n} = a + (n - 1)d]$, we get
\[19 = 1 + (n - 1) \times 2\]
Solving the equation mathematically, we get
\[19 = 1 + 2n - 2\]
\[\Rightarrow 19 = 2n - 1\]
Hence, total terms in the respective series is,
\[2n = 20\]
\[\Rightarrow n = 10\]
Now, hence the sum for the respective series $\left( {1 + 3 + 5 + ..... + 17 + 19} \right)$ can get by using the known formula i.e. ${S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]$
As a result, substituting all the respective values in the formula for $n = 10$, we get
${S_{10}} = \dfrac{{10}}{2}\left[ {2\left( 1 \right) + (10 - 1)2} \right]$
Solving the equation mathematically, we get
${S_{10}} = 5\left[ {2 + 9 \times 2} \right]$
$\Rightarrow {S_{10}} = 5\left( {20} \right)$
Hence, the sum for the respective sequence of all the terms is,
${\left( {{S_{10}}} \right)_1} = 100$ … (ii)
Similarly,
Considering the first sequence i.e. $\left( {2 + 4 + 6 + 0.... + 18 + 20} \right)$
It seems that, $[{t_n} = a + (n - 1)d]$ represents total terms in the respective sequence
Where,
‘${t_n}$’ is the last term of the arithmetic series, i.e. ${t_n} = 20$
‘a’ is the first term of the sequence or series, i.e. $a = 2$
‘n’ is the terms in the respective series or sequence,
‘d’ is the common difference of the arithmetic series expressed as $[d = {t_2} - {t_1}]$ where, ${t_2}$is second term and \[{t_1}\] is first term of the series respectively, i.e. $d = {t_2} - {t_1} = 4 - 2 = 2$
Now, substituting the values in the equation $[{t_n} = a + (n - 1)d]$, we get
\[20 = 2 + (n - 1) \times 2\]
Solving the equation mathematically, we get
\[20 = 2 + 2n - 2\]
\[\Rightarrow 20 = 2n\]
Hence, total terms in the respective series is,
\[n = 10\]
Now, hence the sum for the respective series $\left( {2 + 4 + 6 + 0.... + 18 + 20} \right)$ can get by using the known formula i.e. ${S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]$
As a result, substituting all the respective values in the formula for $n = 10$, we get
${S_{10}} = \dfrac{{10}}{2}\left[ {2\left( 2 \right) + (10 - 1)2} \right]$
Solving the equation mathematically, we get
${S_{10}} = 5\left[ {4 + 9 \times 2} \right]$
$\Rightarrow {S_{10}} = 5\left( {22} \right)$
Hence, the sum for the respective sequence of all the terms is,
${\left( {{S_{10}}} \right)_2} = 110$ … (iii)
Now,
From (ii) and (iii),
It seems that the total sum of the entire sequence is,
$Total{\text{ s}}um = - \left( {1 + 3 + 5 + ..... + 17 + 19} \right) + \left( {2 + 4 + 6 + ..... + 18 + 20} \right)$
\[Total{\text{ s}}um = - {\left( {{S_{10}}} \right)_1} + {\left( {{S_{10}}} \right)_2}\]
Substituting the values we get,
$Total{\text{ s}}um = - 100 + 110$
$\Rightarrow Total{\text{ s}}um = 10$
$\therefore $The option (b) is absolutely correct.

Note: Remember the formulae for ${t_n} = a + (n - 1)d$ which gives the exact term in the sequence of Arithmetic Progression (AP). We can also take the sum of all the terms of the AP sequence by using the formula ${S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]$respectively. One must keenly observe the desired sequence of the AP to determine the respective solution.