The sum of all the even integers between 99 and 301 is 20200
To find the sum of even integers between 99 and 301, we will use the arithmetic progressions(AP). The even numbers can be considered as an AP with common difference 2.
In this case, the first even integer will be 100 and the last even integer will be 300.
nth term of the AP = first term + (n-1) x common difference
⇒ 300 = 100 + (n-1) x 2
Therefore, n = (200 + 2 )/2 = 101
That is, there are 101 even integers between 99 and 301.
Sum of the 'n' terms in an AP = n/2 ( first term + last term)
= 101/2 (300+100)
= 20200
Thus sum of all the even integers between 99 and 301 = 20200
Learn more about arithmetic progressions at brainly.com/question/24592110
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Yes I work this out so yea you divide idk
Answer:
y = 4
Step-by-step explanation:
I assume you want to find the slope-intercept form of the given information.
We are given the slope and a point, so we can find the y-intercept.
y = 0x + b
4 = 0(-2) + b
4 = 0 + b
4 = b
Put everything we know/solved for back into the formula [ y = mx + b ]
y = 4
Best of Luck!
Answer:
0.064
Step-by-step explanation:
solve the btackets first then the powers then start multiplying
