find a formula for the general term of the following sequences 14,11, 8,5,2​

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1. Answer:

   Step by Step Solution

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   14,11,8,5

   Your input 14,11,8,5 appears to be an arithmetic sequence

   Find the difference between the members

   a2-a1=11-14=-3

   a3-a2=8-11=-3

   a4-a3=5-8=-3

   The difference between every two adjacent members of the series is constant and equal to -3

   General Form: a

   n

   =a

   1

   +(n-1)d

   a

   n

   =14+(n-1)*(-3)

   a1=14 (this is the 1st member)

   an=5 (this is the last/nth member)

   d=-3 (this is the difference between consecutive members)

   n=4 (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:

   14+11+8+5

   This sum can be found quickly by taking the number n of terms being added (here 4), multiplying by the sum of the first and last number in the progression (here 14 + 5 = 19), and dividing by 2:

   n(a1+an)

   2

   4(14+5)

   2

   The sum of the 4 members of this series is 38

   This series corresponds to the following straight line y=-3x+14

   Finding the n

   th

   element

   a1 =a1+(n-1)*d =14+(1-1)*-3 =14

   a2 =a1+(n-1)*d =14+(2-1)*-3 =11

   a3 =a1+(n-1)*d =14+(3-1)*-3 =8

   a4 =a1+(n-1)*d =14+(4-1)*-3 =5

   a5 =a1+(n-1)*d =14+(5-1)*-3 =2

   a6 =a1+(n-1)*d =14+(6-1)*-3 =-1

   a7 =a1+(n-1)*d =14+(7-1)*-3 =-4

   a8 =a1+(n-1)*d =14+(8-1)*-3 =-7

   a9 =a1+(n-1)*d =14+(9-1)*-3 =-10

   a10 =a1+(n-1)*d =14+(10-1)*-3 =-13

   a11 =a1+(n-1)*d =14+(11-1)*-3 =-16

   a12 =a1+(n-1)*d =14+(12-1)*-3 =-19

   a13 =a1+(n-1)*d =14+(13-1)*-3 =-22

   a14 =a1+(n-1)*d =14+(14-1)*-3 =-25

   a15 =a1+(n-1)*d =14+(15-1)*-3 =-28

   a16 =a1+(n-1)*d =14+(16-1)*-3 =-31

   a17 =a1+(n-1)*d =14+(17-1)*-3 =-34

   a18 =a1+(n-1)*d =14+(18-1)*-3 =-37

   a19 =a1+(n-1)*d =14+(19-1)*-3 =-40

   a20 =a1+(n-1)*d =14+(20-1)*-3 =-43

   a21 =a1+(n-1)*d =14+(21-1)*-3 =-46

   a22 =a1+(n-1)*d =14+(22-1)*-3 =-49

   a23 =a1+(n-1)*d =14+(23-1)*-3 =-52

   a24 =a1+(n-1)*d =14+(24-1)*-3 =-55

   a25 =a1+(n-1)*d =14+(25-1)*-3 =-58

   a26 =a1+(n-1)*d =14+(26-1)*-3 =-61

   a27 =a1+(n-1)*d =14+(27-1)*-3 =-64

   a28 =a1+(n-1)*d =14+(28-1)*-3 =-67

   a29 =a1+(n-1)*d =14+(29-1)*-3 =-70

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