\( x ; 4 ; y \) form an arithmetic sequence while \( x ; \sqrt{15} ; y \) form a geometric sequence. Calculate \( x \) and \( y \). Consider an arithmetic series: \( S_{n}=n^{2}+3 n \) : Calculate the second term of this series. The first two terms of this sequence form a geometric sequence. Calculate the sum of the first 6 terms of this geometric sequence.

1. We are given that - \(x,\,4,\,y\) form an arithmetic sequence. In an arithmetic sequence the difference between consecutive terms is constant, so \[ 4-x = y-4. \] This implies \[ y = 8-x. \] 2. We are also given that - \(x,\,\sqrt{15},\,y\) form a geometric sequence. In a geometric sequence the ratio between consecutive terms is constant, so \[ \frac{\sqrt{15}}{x} = \frac{y}{\sqrt{15}}. \] Cross multiplying gives \[ y = \frac{15}{x}. \] 3. Equate the two expressions for \(y\): \[ 8-x = \frac{15}{x}. \] Multiply both sides by \(x\) (assuming \(x\neq0\)): \[ x(8-x)=15. \] Expanding and rearranging: \[ 8x - x^2 = 15 \quad \Longrightarrow \quad x^2-8x+15=0. \] Factor the quadratic: \[ (x-3)(x-5)=0. \] Thus, \[ x=3 \quad\text{or}\quad x=5. \] Then, using \(y=8-x\): - If \(x=3\) then \(y=8-3=5\). - If \(x=5\) then \(y=8-5=3\). Therefore, the pair \((x,y)\) can be \((3,5)\) or \((5,3)\). 4. Next, we consider the arithmetic series with the sum of the first \(n\) terms given by \[ S_{n}=n^{2}+3n. \] The \(n\)th term \(a_n\) of an arithmetic series is found by \[ a_n=S_n-S_{n-1}. \] To find the second term \(a_2\), compute: \[ S_{2}=2^{2}+3\cdot2=4+6=10, \] \[ S_{1}=1^{2}+3\cdot1=1+3=4. \] Therefore, \[ a_{2}=S_{2}-S_{1}=10-4=6. \] 5. The first two terms of this sequence (the arithmetic series) are \(a_{1}\) and \(a_{2}\). We already have: \[ a_{1}=S_{1}=4 \quad\text{and}\quad a_{2}=6. \] They form the first two terms of a geometric sequence. The common ratio \(r\) is: \[ r=\frac{a_{2}}{a_{1}}=\frac{6}{4}=\frac{3}{2}. \] 6. We wish to calculate the sum of the first 6 terms of this geometric sequence. The sum \(G_{n}\) of the first \(n\) terms of a geometric sequence with first term \(g_{1}\) and common ratio \(r\) (with \(r\neq 1\)) is \[ G_{n}=g_{1}\,\frac{r^n-1}{r-1}. \] For \(n=6\), \(g_{1}=4\) and \(r=\frac{3}{2}\): \[ G_{6}=4\,\frac{\left(\frac{3}{2}\right)^6-1}{\frac{3}{2}-1}. \] First, calculate \(\left(\frac{3}{2}\right)^6\): \[ \left(\frac{3}{2}\right)^6=\frac{3^6}{2^6}=\frac{729}{64}. \] Next, the denominator: \[ \frac{3}{2}-1=\frac{1}{2}. \] Now substitute back: \[ G_{6}=4\,\frac{\frac{729}{64}-1}{\frac{1}{2}}=4\,\frac{\frac{729-64}{64}}{\frac{1}{2}}=4\,\frac{\frac{665}{64}}{\frac{1}{2}}. \] Dividing by \(\frac{1}{2}\) is equivalent to multiplying by 2: \[ G_{6}=4\,\frac{665}{64}\times 2=\frac{4\times2\times665}{64}=\frac{8\times665}{64}. \] Simplify: \[ \frac{8}{64}=\frac{1}{8}, \] so \[ G_{6}=\frac{665}{8}. \] Final Answers: - For the arithmetic and geometric sequences: \(x=3,\, y=5\) or \(x=5,\, y=3\). - The second term \(a_2\) of the arithmetic series is \(6\). - The sum of the first 6 terms of the geometric sequence is \(\frac{665}{8}\).