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Natali5045456 [20]
3 years ago
11

Write the first 6 terms of the arithmetic sequence whose first term is 8 and has a common difference of 2

Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
6 0

Answer: The first 6 terms are = 8, 10, 12,14,16,18

Step-by-step explanation:

The NTH term of an Arithmetic Sequence is given as

an = a1 + (n - 1 ) d

where a1 = First term  given as 8 and

d=  common difference given as 2

Therefore  We have that

the first term

an = a1 + (n - 1 ) d = 8+(1-1) 2

a1= 8

second term=

an = a1 + (n - 1 ) d= a2= 8 + (2-1) 2

= 8+ 2(1) = 10

3rd term

an = a1 + (n - 1 ) d= a3= 8 + (3-1) 2

= 8+ 2(2)= 8 + 4=12

4th term

an = a1 + (n - 1 ) d= a4= 8 + (4-1) 2

= 8+ 2(3)= 8+6=14

5th term

an = a1 + (n - 1 ) d= a5= 8 + (5-1) 2

= 8+ 2(4)=8+ 8=16

6th term

an = a1 + (n - 1 ) d= a6= 8 + (6-1) 2

= 8+ 2(5)=8 +10 =18

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Please help me, thank you!
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a. The first four terms are -4 , -4/3, -4/9 , -4/27

b. The series is converge

c. The series has sum to ∞ , the sum of the series is -6

Step-by-step explanation:

* Lets revise the geometric series

- Geometric series:

- There is a constant ratio between each two consecutive numbers

- Ex:

# 5  ,  10  ,  20  ,  40  ,  80  ,  ………………………. (×2)

# 5000  ,  1000  ,  200  ,  40  ,  …………………………(÷5)

* General term (nth term) of a Geometric Progression:

- U1 = a  ,  U2  = ar  ,  U3  = ar2  ,  U4 = ar3  ,  U5 = ar4

- Un = ar^n-1, where a is the first term , r is the constant ratio

 between each two consecutive terms  and n is the position of the

  number in the sequence

* In the problem

∵ The Un = -4(1/3)^n-1

∴ a = -4

∴ r = 1/3

a) To find the first four numbers use n = 1, 2 , 3 , 4

∴ U1 = a = -4

∴ U2 = -4(1/3)^(2 - 1) = -4(1/3) = -4/3

∴ U3 = -4(1/3)^(3 - 1) = -4(1/3)^2 = -4(1/9) = -4/9

∴ U4 = -4(1/3)^(4 - 1) = -4(1/3)^3 = -4(1/27) = -4/27

* The first four terms are -4 , -4/3, -4/9 , -4/27

b) If IrI < 1  then the geometric series is converge and if IrI > 1

   then the geometric series is diverge

∵ r = 1/3

∴ The series is converge  

c. The convergent series has sum to ∞

- The rule is: S∞ = a/(1 - r)

∴ S∞ = -4/(1 - 1/3) = -4/(2/3) = -4 × 3/2 = -6

* The sum of the series is -6

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3 years ago
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