Fastest way to calculate the following equations

[Kyoma]

Homework Statement

Find n.

5a + 10d = 26

5a + 5/2(n-1)d + 5/2(n-5)d = 726

n/2(2a + (n-1)d) = 10904

Note that a and d are unknowns as well.

2. The attempt at a solution

It is extremely tedious to calculate by hand. Furthermore, my calculator is only able to solve both linear equations and quadratic equations in the form of ax + b and ax^2 + bx + c respectively, where a, b and c are known constants. However, the above equations give me unknowns like nd, na, which I am unable to key it into the calculator. Hence, is there a better way to solve the above equations? Much help appreciated. Thanks!

 

[mfb]

If you have access to a computer, use that. WolframAlpha can solve it directly with your equations.

Without a computer: I would simplify the second equation to
26-10d + 5d/2 (2n-6) = 726 (using the first equation to get rid of a) and then
-2d + d (n-3)=140
d(n-5)=140
Therefore, d=140/(n-5)

Using this in the last equation gives a cubic equation -> bad. If d is an integer, n has to be at least 145. As 145/2*145 is of the order of 10904, I would try this first. n=145, d=1, a=16/5 leads to a solution, and I think the derivative of the last equation with respect to n will show that this is the only solution.

 

[Ray Vickson]

If you have access to a computer, use that. WolframAlpha can solve it directly with your equations.

Without a computer: I would simplify the second equation to
26-10d + 5d/2 (2n-6) = 726 (using the first equation to get rid of a) and then
-2d + d (n-3)=140
d(n-5)=140
Therefore, d=140/(n-5)

Using this in the last equation gives a cubic equation -> bad. If d is an integer, n has to be at least 145. As 145/2*145 is of the order of 10904, I would try this first. n=145, d=1, a=16/5 leads to a solution, and I think the derivative of the last equation with respect to n will show that this is the only solution.

You can get a quadratic equation in n, by solving the first eqn for a in terms of d, substituting that into the second equation to get a simple linear equation for d in in terms of n. Then, substituting both into the third equation gives a quadratic in n.

 

[mfb]

Oh right, it is just a quadratic equation.
An ugly one, however.

 

[Ray Vickson]

Oh right, it is just a quadratic equation.
An ugly one, however.

I think I must have made a typo when entering the equations before; when I (or Maple) do it again I obtain a very simple linear equation for n (not a quadratic). Solving the first two equations for 'a' and 'd' in terms of n, we have
d = 140/(n-5), a = 2/5*(13*n-765)/(n-5). Putting these into the third equation gives something surprisingly simple.