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garik1379 [7]
3 years ago
12

find the sum of even numbers between 1001 and 10,000 (Arithmetic progressions) DO STEP BY STEP PLEASE ​

Mathematics
1 answer:
love history [14]3 years ago
7 0

Step-by-step explanation:

arithmetic progression

even numbers between 1001 and 10000

1002,1004,.......9998

Tn=a+(n-1)d

Tn = 9998 last term in the progression of even numbers

a = 1002 first term

d = 2 distance

9998=1002+(n-1)(2)

8996=2n-2

8998=2n

n = 4499 the number of terms

sum = n/2 (a+Tn) = 4499/2(1002+9998)

= (4499/2)(11000) = 24744500

Logic:

notice what happens when you add the first and last terms

1002+9998=11000

1004+9996=11000

.......

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2(m-3)+7<21 PLEASE HELP!!!
Contact [7]

Simplify. Distribute 2 to all terms within the parenthesis

2(m -3) = 2m - 6

Simplify. Combine like terms

2m - 6 + 7 < 21

2m + 1 < 21

Isolate the m. Subtract 1 from both sides

2m + 1 (-1) < 21 (-1)

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Isolate the m. Divide 2 from both sides

2m/2 < 20/2

m < 20/2

m < 10

m < 10 is your answer

hope this helps

4 0
3 years ago
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