Mathematics, 15.10.2019 08:30 addisonrausch

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

Answers: 3

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Answers

Answer from: torresnoemi899

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 moreGeometric Series : linkArithmetic Progression : linkGeometric Sequence : linkAnswer details

Grade: Middle School

Subject: Mathematics

Chapter: Arithmetic and Geometric Series

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

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

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Tentukan hasil dari (tanpa menghitung satu persatu)
a. 1+3+5+7+9++99
b. 1-2+3-4+5-6+7-8+...

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