# If o1, o2, on, angles formed around the point o and m(o1) = n degrees, m(o2) = n+1 degrees...m(on) = 2n-1 degrees , m(on + 1) = 2n degrees .Which is the value of n , natural number ?

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Since o1,o2,...,on are angles around a point:

Then the sum of the measures of these angles is 360 degrees;

==> m(o1)+m(o2)+...+m(on)= 360

==> n + n+1 + n+2+ ...+2n-3+2n-2+2n-1 = 360

Rearrange:

(1+2+3+...+n-1 +n )+ n(n+1)=360

we know that: 1+2+3+...+n-1+n = n(n+1)/2

==> n(n+1)/2 +n(n+1)=360

n^2+n +2n^2+2n= 720

3n^2+3n=720

3n(n+1)=720

n(n+1)= 240

n^2+n-240=0

(n+16)(n-15)=0

==> n =15 degrees.

(we will not consider n=-16 because it is a negative value and degrees should be positive values)

The sum of the angles angles give are :

Sum{ m(0i) , i =1 to n} =n+ (n+1)+(n+2)+(n+3) +..... (n+n-1) +(n+n-1) +(n+n). there are n+1 terms.

= n*(n+1)+ [0+1+2+3+...n-1+n]. But since 1+2+3+..+k = k*(k+1)/2

= n(n+1)+n(n+1)/2

= n(n+1)(1+1/2) = n(n+1)(1.5) but this is equal to 2pi or 360 degrees. So n(n+1) = 360/1.5 = 240. Or 15*16 = 240. So there are 15 angles.

Given the measures of the angles and knowing the fact that there are the angles formed around a common point O, we could find the sum of these angles:

m (O1) + m (O2) +...+ m (On) = 360 degrees

Now we'll substitute the terms from frelation above by their values from enunciation.

n + n+1 + n+2 + ... + 2n-1 + 2n = 360 degrees

We could use a trick and write 2n = n+n

n + n+1 + n+2 + ... + (n + n -1) + (n+n)= 360 degrees

Re-arranging the terms above:

n(n+1) + 1 + 2 + 3 +...+(n-1) + n = 360 degrees

But 1 + 2 + 3 +...+(n-1) + n is the addition of the first n natural numbers and the sum of them is n(n+1)/2.

n(n+1) + n(n+1)/2 = 360 degrees

We'll factorize and we'll get:

n(n+1)(1 + 1/2)= 360 degree

n(n+1)3/2= 360 degree

3n(n+1) = 2*360 degree

n(n+1) = 2*120

n(n+1) = 2*12*10

n(n+1) = 2*6*2*10

n(n+1) = 2*2*3*2*2*5

n(n+1) = 2^4*3*5

n(n+1)=16*15

But 16*15=15*16

So, we'll re-write

n(n+1) = 15*16

**n=15 degrees**