answer:Since the common ratio q = frac {1}{2}, according to the sum formula of a geometric progression, we have: S_n = a_1 frac{1 - q^n}{1 - q} where a_1 is the first term of the sequence. For n = 4, the sum of the first four terms is: S_4 = a_1 frac{1 - left(frac{1}{2}right)^4}{1 - frac{1}{2}} The third term of the sequence, a_3, can be expressed as: a_3 = a_1 q^2 = a_1 left(frac{1}{2}right)^2 = frac{a_1}{4} Now let's calculate the desired ratio: frac {S_{4}}{a_{3}} = frac {a_1 frac{1 - left(frac{1}{2}right)^4}{1 - frac{1}{2}}}{frac{a_1}{4}} We simplify this expression by multiplying both the numerator and denominator by 4: frac {S_{4}}{a_{3}} = frac {4 left(1 - left(frac{1}{2}right)^4right)}{1 - frac{1}{2}} We can further simplify it by calculating the expression inside the parentheses and by performing the division in the denominator: frac {S_{4}}{a_{3}} = frac {4 left(1 - frac{1}{16}right)}{frac{1}{2}} frac {S_{4}}{a_{3}} = frac {4 left(frac{16}{16} - frac{1}{16}right)}{frac{1}{2}} frac {S_{4}}{a_{3}} = frac {4 times frac{15}{16}}{frac{1}{2}} frac {S_{4}}{a_{3}} = frac {4 times 15}{16 times frac{1}{2}} frac {S_{4}}{a_{3}} = frac {60}{8} frac {S_{4}}{a_{3}} = boxed{frac {15}{2}} So the answer is boxed{frac {15}{2}}.

answer:Let the foci of the ellipse be F_1 = (4, 3) and F_2 = (-6, 10). We are given that for any point P = (x,y) on the ellipse, the sum of the distances from P to each focus is constant and equals 24. The distance between the foci can be calculated using the distance formula: [ F_1F_2 = sqrt{(4+6)^2 + (3-10)^2} ] [ F_1F_2 = sqrt{10^2 + (-7)^2} ] [ F_1F_2 = sqrt{100 + 49} ] [ F_1F_2 = sqrt{149} ] [ F_1F_2 = boxed{sqrt{149}}. ] Conclusion: The distance between the foci of the given ellipse is sqrt{149} units.

answer:First, solve the equations ( 4x^2 + y^2 = 4 ) and ( x^2 - n(y-1)^2 = 1 ) simultaneously. To eliminate ( x^2 ), subtract the second equation from the first: [ 4x^2 + y^2 - (x^2 - n(y-1)^2) = 4 - 1, ] [ 3x^2 + n(y-1)^2 + y^2 = 3. ] Rearrange and combine like terms: [ 3x^2 + (n+1)y^2 - 2ny + n = 3. ] To make the curves tangent, set the discriminant of the quadratic in ( y ) to zero: [ (-2n)^2 - 4(n+1)(n-3) = 0. ] Simplify and solve for ( n ): [ 4n^2 - 4n^2 - 4n + 12 = 0, ] [ -4n + 12 = 0, ] [ n = 3. ] Thus, ( n = boxed{3} ).

answer:Given ( x, y, z in mathbb{R}_1 ) and ( xyz = 1 ), we need to prove: [ frac{x^{3}}{(1+y)(1+z)} + frac{y^{3}}{(1+x)(1+z)} + frac{z^{3}}{(1+x)(1+y)} geqslant frac{3}{4} ] 1. Writing the inequality explicitly and applying the technique: [ frac{x^{3}}{(1+y)(1+z)} + frac{y^{3}}{(1+x)(1+z)} + frac{z^{3}}{(1+x)(1+y)} ] 2. Consider the expression and use the known relationships ( xyz = 1 ): [ frac{x^{3}}{(1+y)(1+z)} + frac{y^{3}}{(1+x)(1+z)} + frac{z^{3}}{(1+x)(1+y)} ] 3. Rewrite each term by multiplying numerator and denominator by respective variables: [ = frac{x^2 cdot x}{(1+y)(1+z)} + frac{y^2 cdot y}{(1+x)(1+z)} + frac{z^2 cdot z}{(1+x)(1+y)} ] 4. Applying Titu's Lemma on the fractions, we get: [ geq frac{left(x^2 + y^2 + z^2right)^2}{x(1+y)(1+z) + y(1+x)(1+z) + z(1+x)(1+y)} ] 5. Simplify the denominator: [ begin{aligned} x(1+y)(1+z) + y(1+x)(1+z) + z(1+x)(1+y) &= x(1 + y + z + yz) + y(1 + x + z + xz) + z(1 + x + y + xy) &= x + xy + xz + xyz & quad + y + yx + yz + yxz & quad + z + zx + zy + zxy &= x + y + z + 2(xy + yz + zx) + xyz &= x + y + z + 2(xy + yz + zx) + 1 end{aligned} ] Since ( xyz = 1 ), the denominator simplifies to: [ x + y + z + 2(xy + yz + zx) + 1 = x + y + z + 2(xy + yz + zx) + 3 ] 6. Using the inequality: ( x leq frac{x^2 + 1}{2} ). This implies: [ x + y + z leq frac{x^2 + y^2 + z^2 + 3}{2} ] Also: [ 2(xy + yz + zx) leq 2(x^2 + y^2 + z^2) ] 7. Combining all parts: [ x + y + z + 2(xy + yz + zx) + 3 leq frac{x^2 + y^2 + z^2 + 3}{2} + 2(x^2 + y^2 + z^2) + 3 = frac{5}{2} (x^2 + y^2 + z^2) + frac{9}{2} ] 8. Substituting back: [ begin{aligned} frac{left(x^2 + y^2 + z^2right)^2}{x + y + z + 2(xy + yz + zx) + 3} &geq frac{left(x^2 + y^2 + z^2right)^2}{frac{5}{2}(x^2 + y^2 + z^2) + frac{9}{2}} &geq frac{left(x^2 + y^2 + z^2right)^2}{frac{5}{2}left(x^2 + y^2 + z^2right) + frac{3}{2}left(x^2 + y^2 + z^2right)} &= frac{left(x^2 + y^2 + z^2right)^2}{4left(x^2 + y^2 + z^2right)} &= frac{left(x^2 + y^2 + z^2right)}{4} ] 9. Given ( x^2 + y^2 + z^2 geqslant 3 ) (by AM-GM inequality or as ( x y z = 1 )), [ frac{x^2 + y^2 + z^2}{4} geq frac{3}{4} ] Overall, we reach: [ boxed{frac{x^3}{(1+y)(1+z)} + frac{y^3}{(1+x)(1+z)} + frac{z^3}{(1+x)(1+y)} geq frac{3}{4}} ]