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6.) 5,10,15,20,___,___,___
RULE:

7.) 3,5,7,9,___,___,___
RULE:

8.) 12,13,14,15,___,___,___
RULE:

9.) 2,5,8,11,___,___,___
RULE:

10.) 3,9,27,___,___,___
RULE:

Sagot :

gatica136507140067

Answer:

6. 5,10,15,20,25,30,35

RULE: +5

7. 3,5,7,9,11,13,15

RULE: +2

8. 12,13,14,15,16,17,18

RULE: +1

9. 2,5,8,11,14,17,20

RULE: +3

10. 3,9,27,81,243,729

RULE: ×3

Nzm

SOLUTION:

6.) 5,10,15,20,___,___,___

Notice that:

  • 5(1) = 5 (First term)
  • 5(2) = 10 (Second term)
  • 5(3) = 15 (Third term)
  • 5(4) = 20 (Fourth term)

We multiply 5 by n to get the value of the nth term.

Hence, the rule is 5n

Now, we can solve for the missing terms since we know the rule.

Fifth term = 5(5) = 25

Sixth term = 5(6) = 30

Seventh term = 5(7) = 35

7.) 3,5,7,9,___,___,___

Notice that:

  • 5 - 3 = 2
  • 7 - 5 = 2
  • 9 - 7 = 2

Since the terms have a common difference of 2, we will use the arithmetic progression formula to solve for nth term

The formula is:

[tex]\text{a}_{\text{n}} = \text{a}_1 + (\text{n} - 1)\text{d}[/tex]

Plug in the value of a₁ and d.

[tex]\text{a}_{\text{n}} = 3 + (\text{n} - 1)2[/tex]

[tex]\text{a}_{\text{n}} = 3 + \text{2n} - 2[/tex]

[tex]\text{a}_{\text{n}} = \boxed{\text{2n} + 1}[/tex]

Now, we can solve for the missing terms since we know the rule.

Fifth term = 2(5) + 1 = 11

Sixth term = 2(6) + 1 = 13

Seventh term = 2(7) + 1 = 15

8.) 12,13,14,15,___,___,___

Notice that:

  • 13 - 12 = 1
  • 14 - 13 = 1
  • 15 - 14 = 1

Since the terms have a common difference of 1, we will also use again the arithmetic progression formula to solve for nth term

[tex]\text{a}_{\text{n}} = \text{a}_1 + (\text{n} - 1)\text{d}[/tex]

[tex]\text{a}_{\text{n}} = 12 + (\text{n} - 1)1[/tex]

[tex]\text{a}_{\text{n}} = 12 + \text{n} - 1[/tex]

[tex]\text{a}_{\text{n}} = \boxed{\text{n} + 11}[/tex]

Now, we can solve for the missing terms since we know the rule.

Fifth term = 5 + 11 = 16

Sixth term = 6 + 11 = 17

Seventh term = 7 + 11 = 18

9.) 2,5,8,11,___,___,___

Notice that:

  • 5 - 2 = 3
  • 8 - 5 = 3
  • 11 - 8 = 3

Again, since this sequence is an ap, we will use the formula we used on #7 and #8.

[tex]\text{a}_{\text{n}} = \text{a}_1 + (\text{n} - 1)\text{d}[/tex]

[tex]\text{a}_{\text{n}} = 2 + (\text{n} - 1)3[/tex]

[tex]\text{a}_{\text{n}} = 2 + 3\text{n} - 3[/tex]

[tex]\text{a}_{\text{n}} = \boxed{3\text{n} - 1}[/tex]

Now, we can solve for the missing terms since we know the rule.

Fifth term = 3(5) - 1 = 14

Sixth term = 3(6) - 1 = 17

Seventh term = 3(7) - 1 = 20

10.) 3,9,27,___,___,___

Notice that:

  • 3¹ = 3 (First term)
  • 3² = 9 (Second term)
  • 3³ = 27 (Third term)

We raise 3 to the n term to get the nth term.

Hence, the rule is 3ⁿ

Now, we can solve for the missing terms since we know the rule.

Fourth term = 3⁴ = 81

Fifth term = 3⁵ = 243

Sixth term = 3⁶ = 729

ANSWERS:

6.) 5,10,15,20,25,30,35

RULE: 5n

7.) 3,5,7,9,11,13,15

RULE: 2n + 1

8.) 12,13,14,15,16,17,18

RULE: n + 11

9.) 2,5,8,11,14,17,20

3n - 1

10.) 3,9,27,81,243,729

RULE: 3ⁿ

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