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Six numbers form an arithmetic sequence with a common difference of 4. The sum of these numbers equals to 12. What are the six numbers?

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Sagot :

Answer:

-8, -4, 0, 4, 8, 12

Step-by-step explanation:

The nth term Tn of  an arithmetic sequence with a common difference d and a first term a is given as

Tn = a + (n - 1) d

Hence if the first term is a and the common difference is d then

the

2nd term = a + 4

3rd term = a + (3 - 1) 4 = a + 8

4th term = a + (4 - 1) 4 = a + 12

5th term = a + (5 - 1) 4 = a + 16

6th term = a + (6 - 1) 4 = a + 20

If the sum of these six numbers is 12 then

a + a + 4 + a + 8 + a + 12 + a + 16 + a + 20 = 12

6a + 60 = 12

collect like terms

6a = 12 - 60

6a = -48

divide both sides by 6

a= -8

the numbers are

a = -8

2nd term = a + 4 = -8 + 4 = -4

3rd term = a + (3 - 1) 4 = a + 8 = -8 + 8 = 0

4th term = a + (4 - 1) 4 = a + 12 = -8 + 12 = 4

5th term = a + (5 - 1) 4 = a + 16 -8 + 16 = 8

6th term = a + (6 - 1) 4 = a + 20  = -8 + 20 = 12

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