CPhill

Very nice, Max  ....this one is  somewhat difficult  !!!

Could you explain how the area is  8x / 2  ???......my small mind doesn't understand (LOL!!!)

The nth term  = 1 + 5(n  -1)

The sum of the series =  [ first term + last term ] *  number of terms /  2

We can solve this

[  1  +  1 + 5(n - 1) ] (n / 2)  >   40

[ -3 + 5n ] (n)  >  80 

5n^2 - 3n - 80  >  0        set   =   0

5n^2  -3n - 80  = 0

We need n to be  an  integer

Use the Quad Formula to find n and  take its ceiling

n  =  ceiling  (  [ 3 + sqrt [ 3^2  - 4 * 5 * -80 ] ] / 10 )   =   5

The number that is  called is    1 + 5(n -1)  = 1 + 5 (5 - 1)  =  1 + 5 (4)   =  21          

a2 =  a1 + d

a3 = a1 + 2d

So

a1 + (a1 + d)  =  4      →     2a1 + d =  4            (1)

(a1 + d)  + (a1 + 2d)  =  5   →   2a1 + 3d =  5    (2)

Multiply  (1)  by   -3 and  add to (2)

-4a1  =  -7

a1  = -7 / -4  =  7 / 4

S1  = 5

S2 =  15

Then   

a1 =  5     and   a2  = 10         d =  5

So

a15 =  a1   + (15 - 1) (d)  =    5 + (15 - 1) (5)  =  5 + 14  * 5  =  75

S15  =[ a1 + a15 ] ( 15 / 2)  =    [  5 + 75 ] (15 / 2)  =     600

The  slant height of the cone =  21

The circumference of the  cone  =  (120 / 360)  * 2 pi * 21  =  14 pi

The  radius of the cone can  be  found as

14 pi  = 2 pi * r

r = 7  

The height of the cone =  sqrt [ l^2 - r^2 ]  = sqrt [21^2 - 7^2 ]

Volume of the  cone  is

(1/3) pi *  r^2 * h   =

(1/3) pi  * (7)^2 * sqrt [ 21^2 - 7^2 ]  ≈  1015.94

I don't think this is possible  if AB  =  10   ( it must be greater)

If we can....let us imagine this  figure ....  { AB  is  actually ≈ 13.45 }

Let DC = DE =  4

Let AB  = "10"

Let AE =  "x"

And BE  =  BC  =   "10 - x"

And AD  =   sqrt [ DE^2 + AE^2]  = sqrt [ 4^2 + x^2 ] 

Note that ...Triangle ADE ≈  Triangle ABC

So

DE / AD  =  BC / AB

4 / sqrt [ 4^2 + x^2  ] =  (10 - x )  /  10        

The only real solution to  this  is  x =   0  =   AE    which is  impossible

Angle  ABC  =105

Let AD = x    CD = 10 - x

BC / sin 30  = AC /sin 105

BC /sin30 = 10 /sin105

BC = 5/sin105

BA / sin 45 =  AC / sin 105

BA / sin45 = 10 /sin105

BA = (10/sqrt 2) / sin 105  =   20 / (1 + sqrt 3)

Since BD is a  bisector, then

BA / AD = BC / CD

(10/sqrt2)/ sin 105 / x  = 5/sin105 / (10-x)

(10/sqrt 2) (10-x) / sin 105  =  5x / sin105

(10/sqrt 2) (10 -x)  = 5x

100/sqrt 2  - (10/sqrt 2)x = 5x

100/sqrt 2  = (5 + 10/sqrt 2) x

x = (100/sqrt 2) / ( 5 + 10/sqrt 2)  =   (100) / (5sqrt 2 + 10)  =  20 / (2 + sqrt 2) = AD

[ABD ] = (1/2)BA * AD  * sin 30  =  (1/4) [20 / (1 + sqrt 3)] * [ 20 /(2 + sqrt 2) ]  ≈ 10.72

https://web2.0calc.com/questions/triangles_143

It's still a pretty good problem.....!!!

I think it must be  3x^3 - 20x^2 + kx + 12   ????

3 is one  root

3  [  3     -  20           k        12    ]

                   9        -33       3k - 99

    ____________________________

       3       -11      k - 33     3k - 87

3k - 87 =  0

3k =  87

k = 29

3  [  3      -20         29       12  ]   

                  9         -33     -12

   _________________________

      3        -11         -4         0

The remaining polynomial is

3x^2  -11x  - 4   = 0

(3x  + 1) ( x - 4)  = 0

The other two roots   are   -1/3   and   4