Computer Science, asked by fideliskarash01, 1 day ago

Find the sum of the first 12 terms of the series 5, 9, 13, 17,

Answers

Answered by vrickshik
1

Answer: Step by Step Solution

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5,9,13,17,21

Your input 5,9,13,17,21 appears to be an arithmetic sequence

Find the difference between the members

a2-a1=9-5=4

a3-a2=13-9=4

a4-a3=17-13=4

a5-a4=21-17=4

The difference between every two adjacent members of the series is constant and equal to 4

General Form: a

n

=a

1

+(n-1)d

a

n

=5+(n-1)4

a1=5   (this is the 1st member)

an=21  (this is the last/nth member)

d=4  (this is the difference between consecutive members)

n=5  (this is the number of members)

Sum of finite series members

The sum of the members of a finite arithmetic progression is called an arithmetic series.

Using our example, consider the sum:

5+9+13+17+21

This sum can be found quickly by taking the number n of terms being added (here 5), multiplying by the sum of the first and last number in the progression (here 5 + 21 = 26), and dividing by 2:

n(a1+an)

2

5(5+21)

    2

The sum of the 5 members of this series is 65

This series corresponds to the following straight line y=4x+5

Finding the n

th

element

a1 =a1+(n-1)*d =5+(1-1)*4 =5

a2 =a1+(n-1)*d =5+(2-1)*4 =9

a3 =a1+(n-1)*d =5+(3-1)*4 =13

a4 =a1+(n-1)*d =5+(4-1)*4 =17

a5 =a1+(n-1)*d =5+(5-1)*4 =21

a6 =a1+(n-1)*d =5+(6-1)*4 =25

a7 =a1+(n-1)*d =5+(7-1)*4 =29

a8 =a1+(n-1)*d =5+(8-1)*4 =33

a9 =a1+(n-1)*d =5+(9-1)*4 =37

a10 =a1+(n-1)*d =5+(10-1)*4 =41

a11 =a1+(n-1)*d =5+(11-1)*4 =45

a12 =a1+(n-1)*d =5+(12-1)*4 =49

a13 =a1+(n-1)*d =5+(13-1)*4 =53

a14 =a1+(n-1)*d =5+(14-1)*4 =57

a15 =a1+(n-1)*d =5+(15-1)*4 =61

a16 =a1+(n-1)*d =5+(16-1)*4 =65

a17 =a1+(n-1)*d =5+(17-1)*4 =69

a18 =a1+(n-1)*d =5+(18-1)*4 =73

a19 =a1+(n-1)*d =5+(19-1)*4 =77

a20 =a1+(n-1)*d =5+(20-1)*4 =81

a21 =a1+(n-1)*d =5+(21-1)*4 =85

a22 =a1+(n-1)*d =5+(22-1)*4 =89

a23 =a1+(n-1)*d =5+(23-1)*4 =93

a24 =a1+(n-1)*d =5+(24-1)*4 =97

a25 =a1+(n-1)*d =5+(25-1)*4 =101

a26 =a1+(n-1)*d =5+(26-1)*4 =105

a27 =a1+(n-1)*d =5+(27-1)*4 =109

a28 =a1+(n-1)*d =5+(28-1)*4 =113

a29 =a1+(n-1)*d =5+(29-1)*4 =117

a30 =a1+(n-1)*d =5+(30-1)*4 =121

Explanation: Mark me as brainliest pls i found this answer about 1 hrs pls

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