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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The answer would be 3n^2 + 2.
This can be found/proven by replacing "n" with term number (1,2,3,4...), then solving to get the final number. For example 3 * 1^2 + 2. You would first do 1^2, which is 1. Next, you would multiply 1 by 3, to get 3. Finally, you'd and the 2 to get 5. 5 is the 1st term, and you can use this same equation to get the rest of the terms you need.
I hope this helps!
Answer:
<em>AB = 7.35 cm</em>
Step-by-step explanation:
From the attachment,
In ΔDEF,
DF = GH-(GD+FH) = 6 - (2+3) = 1 cm
DE = 2+3 = 5 cm (sum of two radius)
Applying Pythagoras theorem,
In ΔCDI,
DI = GH-(GD+IH) = 6 - (2+1.5) = 2.5 cm
CD = 2+1.5 = 3.5 cm (sum of two radius)
Applying Pythagoras theorem,
AB = EF + CI = 
Answer:
Gain: (Decrease)
Loss:
(Increase)
Step-by-step explanation:
Answer:
is not; 90; right; wrong; 11; wrong
Step-by-step explanation:
