Answer:
The total number of marbles in the set is 107 marbles
Step-by-step explanation:
The parameters given are;
Let the size of the marbles be 1 unit
The marbles are arranged to form the equilateral triangle as follows;
1st row = 1 marble
2nd row = 2 marbles and so on so that the total number of marbles for an arithmetic series as follows;
Total number of marbles, t = 1 + 2 + 3 + ... + n
Therefore;
![t = \dfrac{1 + n}{2} \times n](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7B1%20%2B%20n%7D%7B2%7D%20%5Ctimes%20n)
Since there are 2 marbles left in forming the first equilateral triangle, we have the total number of marbles in the set, t₁, is given as follows;
![t_1 = \dfrac{1 + n}{2} \times n + 2](https://tex.z-dn.net/?f=t_1%20%3D%20%5Cdfrac%7B1%20%2B%20n%7D%7B2%7D%20%5Ctimes%20n%20%2B%202)
When the same marble set are arranged into a triangle in which each side has one more marble than in the first arrangement, there where 13 marble shortage, hence, the total number of marbles is given as follows;
![t_1 = \dfrac{1 + (n+1)}{2} \times (n+1) -13](https://tex.z-dn.net/?f=t_1%20%3D%20%5Cdfrac%7B1%20%2B%20%28n%2B1%29%7D%7B2%7D%20%5Ctimes%20%28n%2B1%29%20-13)
We therefore have;
![t_1 = \dfrac{1 + n}{2} \times n + 2 = \dfrac{1 + (n+1)}{2} \times (n+1) -13](https://tex.z-dn.net/?f=t_1%20%3D%20%20%5Cdfrac%7B1%20%2B%20n%7D%7B2%7D%20%5Ctimes%20n%20%2B%202%20%3D%20%5Cdfrac%7B1%20%2B%20%28n%2B1%29%7D%7B2%7D%20%5Ctimes%20%28n%2B1%29%20-13)
Which gives;
![\dfrac{n^{2}+n+4}{2}=\dfrac{n^{2}+3\cdot n-24}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7Bn%5E%7B2%7D%2Bn%2B4%7D%7B2%7D%3D%5Cdfrac%7Bn%5E%7B2%7D%2B3%5Ccdot%20n-24%7D%7B2%7D)
Therefore;
n² + n + 4 = n² + 3·n - 24
2·n = 24 + 4 = 28
n = 14
From which we have;
![t_1 = \dfrac{1 + 14}{2} \times 14 + 2 = 107](https://tex.z-dn.net/?f=t_1%20%3D%20%5Cdfrac%7B1%20%2B%2014%7D%7B2%7D%20%5Ctimes%2014%20%2B%202%20%3D%20107)
Therefore, the total number of marbles in the set, t₁ = 107 marbles.