Question

Tentukan hasil dari (tanpa menghitung satu persatu)
a. 1+3+5+7+9+.....+99
b. 1-2+3-4+5-6+7-8+.....-100
c. -100-99-98-..........-2-1-0+1+2+.....+48+49+50

Answer 1

a . 1 + 3 + 5 + 7 + 9 + ... + 99 = 2500

b. 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + ... - 100 = -50

c. -100 - 99 - 98 - .... -2 - 1 - 0 + 1 + 2 + ... + 48 + 49 + 50 = -3775

Further explanation

Let us learn about Arithmetic Progression.

Arithmetic Progression is a sequence of numbers in which each of adjacent numbers have a constant difference.

$\large {\boxed {T_n = a + (n-1)d } }$

$\large {\boxed {S_n = \frac{1}{2}n ( 2a + (n-1)d ) } }$

Tn = n-th term of the sequence

Sn = sum of the first n numbers of the sequence

a = the initial term of the sequence

d = common difference between adjacent numbers

Let us now tackle the problem!

Question a :

1 + 3 + 5 + 7 + 9 + ... + 99

initial term = a = 1

common difference = d = ( 3 - 1 ) = 2

Firstly , we will find how many numbers ( n ) in this series.

$T_n = a + (n-1)d$

$99 = 1 + (n-1)2$

$99-1 = (n-1)2$

$98 = (n-1)2$

$\frac{98}{2} = (n-1)$

$49 = (n-1)$

$n = 50$

At last , we could find the sum of the numbers in the series using the above formula.

$S_n = \frac{1}{2}n ( 2a + (n-1)d )$

$S_{50} = \frac{1}{2}(50) ( 2 \times 1 + (50-1) \times 2 )$

$S_{50} = 25 ( 2 + 49 \times 2 )$

$S_{50} = 25 ( 2 + 98 )$

$S_{50} = 25 ( 100 )$

$\large { \boxed { S_{50} = 2500 } }$

Question b :

In this question let us find the series of even numbers first ,  such as :

2 + 4 + 6 + 8 + ... + 100

initial term = a = 2

common difference = d = ( 4 - 2 ) = 2

Firstly , we will find how many numbers ( n ) in this series.

$T_n = a + (n-1)d$

$100 = 2 + (n-1)2$

$100-2 = (n-1)2$

$98 = (n-1)2$

$\frac{98}{2} = (n-1)$

$49 = (n-1)$

$n = 50$

We could find the sum of the numbers in the series using the above formula.

$S_n = \frac{1}{2}n ( 2a + (n-1)d )$

$S_{50} = \frac{1}{2}(50) ( 2 \times 2 + (50-1) \times 2 )$

$S_{50} = 25 ( 4 + 49 \times 2 )$

$S_{50} = 25 ( 4 + 98 )$

$S_{50} = 25 ( 102 )$

$\large { \boxed { S_{50} = 2550 } }$

At last , we could find the result of the series.

1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + ... - 100

= ( 1 + 3 + 5 + 7 + ... + 99 ) - ( 2 + 4 + 6 + 8 + ... + 100 )

= 2500 - 2550

= -50

1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + ... - 100 = -50

Question c :

-100 - 99 - 98 - .... -2 - 1 - 0 + 1 + 2 + ... + 48 + 49 + 50

initial term = a = -100

common difference = d = ( -99 - (-100) ) = 1

Firstly , we will find how many numbers ( n ) in this series.

$T_n = a + (n-1)d$

$50 = -100 + (n-1)1$

$50+100 = (n-1)$

$150 = (n-1)$

$n = 151$

We could find the sum of the numbers in the series using the above formula.

$S_n = \frac{1}{2}n ( 2a + (n-1)d )$

$S_{151} = \frac{1}{2}(151) ( 2 \times (-100) + (151-1) \times 1 )$

$S_{151} = 75.5 ( -200 + 150 )$

$S_{151} = 75.5 ( -50 )$

$\large { \boxed { S_{151} = -3775 } }$

Learn more

• Geometric Series : https://brainly.com/question/4520950
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Answer details

Grade: Middle School

Subject: Mathematics

Chapter: Arithmetic and Geometric Series

Keywords: Arithmetic , Geometric , Series , Sequence , Difference , Term

Answer 2

The sum of the series are:

Part(a): $\fbox{\begin\\\ \math S=2500\\\end{minispace}}$

Part(b): $\fbox{\begin\\\ \math S=-50\\\end{minispace}}$

Part(c): $\fbox{\begin\\\ \math S=-3775\\\end{minispace}}$

Further explanation:

A series is defined as a sum of different numbers in which each term is obtained from a specific rule or pattern.

In this question we need to determine the sum of the series given in the part (a), part (b) and part (c).

Part(a):

The series given in part (a) is as follows:

$1+3+5+7+9+...+99$

All the terms in the given series are odd numbers.

From the given series in part(a) it is observed that the series is an arithmetic series with the common difference of $2$.

An arithmetic series is a series in which each successive member of the series differs from its previous term by a constant quantity.

From the above series it is observed that the first term is $1$, second term is $3$, third term is $5$, fourth term is $7$, fifth term is $9$ and the last term is $99$.

The nth term in a arithmetic series is given as follows:

$a_{n}=a+(n-1)d$           (1)

In the above equation a represents the first term, $n$ represents the total terms and $d$ represents the common difference.

Substitute $99$ for $a_{n}$, $1$ for $a$ and $2$ for $d$ in equation (1).

$\begin{aligned}99&=1+2(n-1)\\2(n-1)&=98\\n-1&=49\\n&=50\end{aligned}$

Therefore, total number of terms in the series is $50$. This implies that $a_{50}=99$.

The sum of an arithmetic series is calculated as follows:

$S_{n}=\dfrac{n}{2}(a+a_{n})$          (2)

Substitute $50$ for $n$ in equation (2).

$\begin{aligned}S_{50}&=\dfrac{50}{2}(a+a_{50})\\&=25\times (1+99)\\&=25\times 100\\&=2500\end{aligned}$

Therefore, the sum of the series for part(a) is $\bf 2500$.

Part(b):

The series given in part (b) is as follows:

$1-2+3-4+5-6+7-8+….-100$

Express the given series as follows:

$S=(1+3+5+7+...+99)-(2+4+6+8+...+100)\\S=S^{'}-S^{''}$

The series $S^{'}$ is as follows:

$S^{'}=1+3+5+7+...+99$

It is observed that the above series $S^{'}$ is exactly same as the series given in the part(a) and the sum of the series of part(a) as calculated above is $2500$.

Therefore, sum of the series $S^{'}$ is $2500$ i.e., $S^{'}=2500$.

The series $S^{"}$ is as follows:

$S^{"}=2+4+6+8+...+100$

From the above series it is observed that the series $S^{"}$ is an arithmetic series as the difference between each consecutive member is $2$ and the last term is $100$.

Substitute $2$ for $a$, $2$ for $d$ and $100$ for $a_{n}$ in equation (1).

$\begin{aligned}100&=2+(n-1)2\\(n-1)2&=98\\n-1&=49\\n&=50\end{aligned}$

This implies that $a_{50}=100$.

To calculate the sum of  substitute $50$ for $n$ in equation (2).

$\begin{aligned}S_{50}&=\dfrac{50}{2}(a+a_{50})\\&=(25)(2+102})\\ &=25\times 102\\&=2550\end{aligned}$

Therefore, sum of the series $S^{"}$ is $2550$.

Substitute $2550$ for $S^{"}$ and $2500$ for  in equation (3).

$\begin{aligned}S&=S^{'}+S^{"}\\&=2500-2550\\&=-50\end{aligned}$

Therefore, the sum of the series for part(b) is $\bf -50$.

Part(c):

The series given in part(c) is as follows:

$-100-99-9-...-2-1-0+1+2+...+48+49+50$

From the above series it is observed that it is an arithmetic series with common difference as $1$, first term as $-100$ and the last term as $50$.

Substitute $-100$ for $a$, $1$ for $d$ and $50$ for $a_{n}$ in equation (1).

$\begin{aligned}50&=-100+(n-1)1\\n-1&=150\\n&=151\end{aligned}$

Substitute $151$ for $n$ in equation (2).

$\begin{aligned}S_{151}&=\dfrac{151}{2}(a+a_{151})\\&=\dfrac{151}{2}(-100+50)\\&=-25\times 151\\&=-3775\end{aligned}$

Therefore, the sum of the series for part(c) is $\bf -3775$.

Learn more:

1. A problem on greatest integer function https://brainly.com/question/8243712

2. A problem to find radius and center of circle https://brainly.com/question/9510228  

3. A problem to determine intercepts of a line https://brainly.com/question/1332667  

Answer details:

Grade: High school

Subject: Mathematics

Chapter: Series

Keywords: Series, sequence, arithmetic sequence, arithmetic series, 1+3+5+7+9+….+99, 1-2+3-4+5-6+7-8+….-100, -100-99-98-….-2-1-0+1+2+…..+48+49+50, sum of series, first term, common difference.