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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Answer:
Step-by-step explanation:
<u>Rule for rounding:</u>
- Round up if number is between 5 and 9
- Round down if number is between 0 and 4
<u>What numbers in hundredths can be rounded to 14.5?</u>
and
<u>From above we see that:</u>
- The greatest decimal is 14.54
- The least decimal is 14.45
Answer:
it is 6 because there is no other explanation to 1-f f=6!
Step-by-step explanation:
all i did was look 1 - f = 5 then f equaled 6!
It is m + 9 because if we let m = 60 and the lest was 9 minutes longer than m, the test would be 69 minutes long. Therefore the answer is m + 9
