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RUDIKE [14]
3 years ago
13

Arithmetic series help!!!!

Mathematics
1 answer:
jekas [21]3 years ago
4 0

The initial term of the sequence is 15. The common difference is 3, so the n-th term (an) is

... an = 15 + 3(n-1)

For n=14, the 14th term of the sequence is

... a14 = 15 + 3(14 -1) = 54

The sum of terms is the average of the first and last, multiplied by the number of terms:

... sum = (15 + 54)/2·14 = 7·69 = 483

The appropriate answer choice is

... 54; 483

You might be interested in
22 less than the quotient of an unknown number and 35 is -25
Alborosie
76-(xdevided5)=116 is the answer

4 0
3 years ago
I'll give you brainliest! Just need help with 5-10! (Least common factors)
umka2103 [35]

Answer:

  • 60
  • 30
  • 48
  • 42
  • 45
  • 55

Step-by-step explanation:

5.

12,\:30\\\\\mathrm{Prime\:factorization\:of\:}12:\quad 2\times\:2\times\:3\\\mathrm{Prime\:factorization\:of\:}30:\quad 2\times\:3\times\:5\\\\=2\times\:2\times\:3\times\:5\\\\=60

6.

6,\:10\\\\\mathrm{Prime\:factorization\:of\:}6:\quad 2\times\:3\\\mathrm{Prime\:factorization\:of\:}10:\quad 2\times \:5\\\\=2\times \:3\times \:5\\\\=30

7.

16,\:24\\\\\mathrm{Prime\:factorization\:of\:}16:\quad 2\times \:2\times\:2\times\:2\\\mathrm{Prime\:factorization\:of\:}24:\quad 2\times\:2\times\:2\times\:3\\\\=2\times\:2\times\:2\times\:2\times\:3\\\\=48

8.

14,21\\\\\mathrm{Prime\:factorization\:of\:}14:\quad 2\times\:7\\\mathrm{Prime\:factorization\:of\:}21:\quad 3\times\:7\\\\=2\times\:7\times\:3\\\\=42

9.

9,15\\\\\mathrm{Prime\:factorization\:of\:}9:\quad 3\times\:3\\\mathrm{Prime\:factorization\:of\:}15:\quad 3\times\:5\\\\=3\times\:3\times\:5\\\\=45

10.

5,\:11\\\\\mathrm{Prime\:factorization\:of\:}5:\quad 5\\\mathrm{Prime\:factorization\:of\:}11:\quad 11\\\\=5\times \:11\\\\=55

5 0
3 years ago
Please help ASAP <br> Questions 1 and 2 plzzzzz
liubo4ka [24]

Take the sequence in 1a

The 10th term is 31

The 20th is 61

If you wanted to find these by continuing the series, you'd have to add 3 to the last number in the series, then 3 more, then 3 more, until you reach the 20th term. By this point, you will have added 3 to the first term 19 times. That's where the formula comes from. So here,

a = 4, the first term

n = 20, the number of the term we need

d = 3, how much we're adding each time between one term and the next

Then, to get the 20th term,

4 + (20 - 1) • 3 = 4 + (19 • 3) = 4 + 57 = 61

Answers

The 10th and 20th terms of each sequences are

a. 31; 61

b. 48; 98

c. 47; 97

(in <em>c</em>, you're adding the same <em>d</em> as in the sequence above, but your first term is one unit less)

d. -25; -75

(same thing as before, but now, <em>d</em> is negative)

e. 11.5; 16.5

(with <em>d</em>=1/2 or 0.5)

f. 6+1/2; 8+1/2

Use these to check your answers after applying the formula, but know that I calculated on the fly and didn't check these.

5 0
3 years ago
Create the pattern with the rule n×3+1
Kitty [74]
3n+1 is the answer
Hope this helps and please mark me as brainlest and like
5 0
3 years ago
What’s is the answer to2(9)+7
AveGali [126]
The answer would be 25

9 times 2 is 18 plus 7 is 25
4 0
3 years ago
Read 2 more answers
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