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

1 Answer

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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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