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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C, 60 hope this helps it helped me on my test
1/10 is the next fraction in the sequence. If you multiply the numerator and denominator by a common factor to give each fraction a denominator of 20, the pattern goes 6/20, 5/20, 4/20, 3/20. The next term would be 2/20 which equals 1/10.
Answer:4
Step-by-step explanation:
2y = 8x
Divide through by 2
2Y/2 = 8X/2
Y = 4X
Therefore the constant of proportionality is 4
Answer:
5=1/3x-11
Step-by-step explanation:
Add eleven to each side to get 16=1/3x. Then multiply each side by three to get 48=x, so x=48.-step explanation: