This is easier than it looks. Imagine finding the sum of the first 8 integers. The obvious way is 1+2+3+4+5+6+7+8. If you slog through this you find the answer is 36. Now let's look at this differently. 1+8=9, 2+7=9, 3+6=9, 4+5=9. There is a pattern here ! All the integers got used, each pair adds up to 9, and there 4 pairs. Four nines make 36, which is the same answer as the other way. So, 1+200=201, 2+199=201, 3+198=201, . . . . 98+103=201, 99+102=201, 100+101=201. All the numbers got used, each pair summed to 201 and there were 100 pairs. 201 X 100 = 20100, and that is the sum of the first 200 integers. Saves a lot of wear and tear on the fingers !
This is easier than it looks. Imagine finding the sum of the first 8 integers. The obvious way is 1+2+3+4+5+6+7+8. If you slog through this you find the answer is 36. Now let's look at this differently. 1+8=9, 2+7=9, 3+6=9, 4+5=9. There is a pattern here ! All the integers got used, each pair adds up to 9, and there 4 pairs. Four nines make 36, which is the same answer as the other way. So, 1+200=201, 2+199=201, 3+198=201, . . . . 98+103=201, 99+102=201, 100+101=201. All the numbers got used, each pair summed to 201 and there were 100 pairs. 201 X 100 = 20100, and that is the sum of the first 200 integers. Saves a lot of wear and tear on the fingers !
"Consecutive" integers are integers that have no other integer between them.
40, 80, 120, 160, 200, 240
Here is an example. You increase something from 200 to 250. First, subtract to get the difference (50). The, divide this difference by the initial number (50 / 200 = 0.25). Finally, convert that into a percentage, by multiplying by 100 (25%).
114 115 116 117 118 119 120 121 122 123 124 125 126
In Order to find the reciporal of any number, you flip the numerator with the denominator, which in this case will give you (1/200)
The sum of the first 200 positive integers is 19,900.
The sum of the first 201 positive integers is 20100 if you include 0 (i.e. from 0 to 200). If you sum the integers from 1 to 201 instead, the sum is 20301.
They are positive integers.
6
Nine of them.
There are 200 positive four digit integers that have 1 as their first digit and 2 or 5 as their last digit. There are 9000 positive four digit numbers, 1000 through 9999. 1000 of them have 1 as the first digit, 1000 through 1999. 200 of them have 2 or 5 as their last digit, 1002, 1005, 1012, 1015, ... 1992, and 1995.
Let the first integer be x. Since the integers are consecutive odds we know that the integers are x, x+2, x+4,and x+6. Since the sum of all of these is 200 we can set up the equation:4x+12 = 200. Solving we get x=47. Therefore the integers are 47, 49, 51, and 53.
use this formula: S=(N/2)(F+L) S= the sum of the first 100 positive even integers N= the number of terms F= first term L=last term Since we are using the first 100 positive integers we will replace "N" with 100. so it will look like this : S=(100/2)(F+L) Next we must substitute the "F" with the first positive term which is the number 2. So now it looks like this: S=(100/2)(2+L) We use the number 2 because our positive integers are 2,4,6,8,10,12...... you get the picture. Now we must substitute the "L" with the last positive term which is 200. As you probably guessed it will look like this: S=(100/2)(2+200) We use the number 200 because it is the last positive even integer. If we used 202 then that would have meant we used the 101th positive even integer and we don't want to do that. == == We should solve the numbers inside the parenthesis first. S=(100/2)(2+200) 100 divided by 2 = 50 So: S=50(2+200) 2+200=202 That means: S=50(202) 50 x 202 = 10,100 Finally: S=10,100 There you go! Wasn't that easy? Note: This formula only works if you are looking for the sum of the first 100 positive or negative, even or odd integers. It sucks cause this formula is the Bomb right? Oh well
8729
The square root of 200 is not an integer (whole number). In rounded form, the two square roots of 200 are positive and negative 14.14214... .
There is no set of two consecutive integers equating to 200.
China is the first documented case in 200 BC