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3 years ago

Thank you to anyone that helps

Mathematics

2 answers:

Margarita [4]3 years ago

7 0

Answer:

Step-by-step explanation:

1 Subtract 150150 from both sides.

10x=200+5x-150

10x=200+5x−150

2 Simplify 200+5x-150200+5x−150 to 5x+505x+50.

10x=5x+50

3 Subtract 5x5x from both sides.

10x-5x=50

10x−5x=50

4 Simplify 10x-5x10x−5x to 5x5x.

5x=50

5 Divide both sides by 55.

x=\frac{50}{5}

6 Simplify \frac{50}{5}

to 1010.

x=10

Done

IceJOKER [234]3 years ago

4 0

I agree with the other person I think that that is the right Answer

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Tentukan hasil dari (tanpa menghitung satu persatu)

liubo4ka [24]

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

<h3>Further explanation</h3>

Let us learn about Arithmetic Progression.

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

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

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

<em>Tn = n-th term of the sequence</em>

<em>Sn = sum of the first n numbers of the sequence</em>

<em>a = the initial term of the sequence</em>

<em>d = common difference between adjacent numbers</em>

Let us now tackle the problem!

<h2>Question a :</h2>

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

<em>initial term = a = 1</em>

<em>common difference = d = ( 3 - 1 ) = 2</em>

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.

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

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

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

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

$S_{50} = 25 ( 100 )$

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

<h2>Question b :</h2>

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

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

<em>initial term = a = 2</em>

<em>common difference = d = ( 4 - 2 ) = 2</em>

<em />

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

<h2>Question c :</h2>

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

<em>initial term = a = -100</em>

<em>common difference = d = ( -99 - (-100) ) = 1</em>

<em />

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.

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

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

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

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

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

<h3>Learn more</h3>

Geometric Series : brainly.com/question/4520950
Arithmetic Progression : brainly.com/question/2966265
Geometric Sequence : brainly.com/question/2166405

<h3>Answer details</h3>

Grade: Middle School

Subject: Mathematics

Chapter: Arithmetic and Geometric Series

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

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