Question 982206
Each tile in the mosaic is in the shape of an equilateral triangle,
{{{12inches=1foot}}} to a side.


On the base of that triangular mosaic,
you have the bases of {{{25}}} tiles of color {{{X}}} .
Tiles of color {{{A}}} will be at all 3 vertices of the triangular mosaic,
and all 3 sides of the triangular mosaic would look just the same,
so it does not matter which one I choose to call the base of the triangular mosaic.


In between the {{{25}}} base-down, vertex-up triangles of color {{{X}}} ,
there will be {{{25-1=24}}} base-up, vertex-down triangles of color {{{Y}}} ,
completing the first row/layer of tiles at the base of the triangular mosaic.


Atop each of those {{{24}}} base-up, vertex-down triangles of color {{{Y}}} ,
there will be one base-down, vertex-up triangle of color {{{X}}} ,
for a total of {{{24}}} base-down, vertex-up triangles of color {{{X}}}
on the second row of the triangular mosaic.


In between the {{{24}}} base-down, vertex-up triangles of color {{{X}}} on the second row,
there will be {{{24-1=23}}} base-up, vertex-down triangles of color {{{Y}}} ,
completing the second row/layer of tiles at the base of the triangular mosaic.


That patter repeats, so you have
{{{N[X]=25+24+23+"..."+3+2+1}}} tiles of color {{{X}}} , and
{{{N[Y]=24+23+22+"..."+3+2+1}} tiles of color {{{Y}}} .
Those numbers are the sums of the {{{25}}} and {{{24}}} first terms of the arithmetic sequence with first term {{{1}}} and common difference {{{1}}} .
The easiest way to calculate the sum {{{a[1]+a[2]+a[3]+"..."+a[n-2]+a[n-1]+a[n]}}} when you know
the number {{{n}}} of terms you are adding,
the first term {{{a[1]}}} you are adding, and
the last term {{{a[n]}}} you are adding is
{{{a[1]+a[2]+a[3]+"..."+a[n-2]+a[n-1]+a[n]=n(a[1]+a[n])/2}}} .
So,
{{{N[Y]=24*(24+1)/2=12*25=highlight(300)}}}
and
{{{N[x]=N[Y]+25=300+25=highlight(325)}}}