find a formula for the general term of each of the following sequences 14;11;8;5;2
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Answer:
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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