Sometime ago, in one of newsletters I came across a question to find next number in series. The series was 3, 11, 20, 27, 29, 23. The next number in series needed to be found. Well, I tried as much as I can and stopped when my brain gave up on me. I was never too bright at math. I wasn’t bad either but I wasn’t this old either at that time.
So can you help me find answer?
A for effort was given to a colleague who answered this correctly and I can clearly see why. He answered it very thoroughly and I am going to publish it with due credits to him (name not published for privacy).
As he wrote:
“The series given was 3, 11, 20, 27, 29, 23
The solution, the way I saw it was to find some function, f(x) that would equal:
f(1) = 3, f(2) = 11, f(3) = 20, etc
so that I could predict the value of f(7).
I found the differences between each set of answers to go down 3 levels, with a shared difference of -3. Three levels means I need to find a third degree polynomial f(x) = ax^3 + bx^2 + cx + d.
I then needed to find the values for a, b, c and d. There are 4 unknowns, so I need 4 equations.
The four equations are:
f(1) = a(1)^3 + b(1)^2 + c(1) + d = 3
f(2) = a(2)^3 + b(2)^2 + c(2) + d = 11
f(3) = a(3)^3 + b(3)^2 + c(3) + d = 20
f(4) = a(4)^3 + b(4)^2 + c(4) + d = 27
Using linear algebra and putting these into a matrix and reducing to echelon form, I solved for a, b, c and d.
a=-1/2, b=7/2, c=1, d=-1
That makes f(x) = -1/2 x^3 + 7/2 x^2 + x – 1
Calculating out f(x):
|X||-1/2 x^3||7/2 x^2||X||-1||f(x)|
Finally, the answer to the question is 6, highlighted above in red.”