What is sum of all odd numbers between 100 and 200

1 answer:

5 0

Odd numbers are of the form:

2n-1

So first we need to know how many odd numbers there are between 100 and 200...

100<2n-1<200 add 1 to all terms

101<2n<201 divide all terms by 2

50.5<n<100.5, since n must be an integer

51≤n≤100 so there are 100-51+1=50 odd numbers

Notice that 2n-1 forms an arithmetic sequence from 101 to 199 in this instance.

The sum of any arithmetic sequence is the average of the first and last terms times the number of terms, mathematically:

s=(a+b)(n/2), where a and b are the first and last terms and n is the number of terms, we saw earlier that there are 50 terms and that a=101 and b=199 so:

s=(101+199)(50/2)

s=7500

So the sum of all odd numbers between 100 and 200 is 7500.

You might be interested in

What is the sum of the squares of the lengths of the $\textbf{medians}$ of a triangle whose side lengths are $10,$ $10,$ and $12

FrozenT [24]

A picture can help.

The median to the long side divides the isosceles triangle into two right triangles with hypotenuse 10 and short leg 6. Thus the long leg (median of interest) is found by the Pythagorean theorem to be

... √(10² -6²) = √64 = 8

Then the midpoint of the short side is found to be 6 + (6/2) = 9 units to the side and 8/2 = 4 units above the opposite vertex. Hence the square of the length of that median is 9² + 4² = 97.

The sum of squares of interest is

... 8² + 2×97 = 258

4 0

3 years ago

Read 2 more answers

WXY is a right triangle. A. True. B. False.

lora16 [44]

Answer:

B. False

Step-by-step explanation:

According to pythagorean theorem, for this to be a right triangle, the sum of square of the length of the two legs must equal square of the length of the hypotenuse (longest side).

So $(\sqrt{34} )^{2}$ should equal $(\sqrt{18} )^{2}+(\sqrt{18} )^{2}$