If the m^{{th}{~}},n^{{th}{~}} and p^{{th}{~}} terms of an A.P and G.P. are equal and are x,y,z then prove that x^{y - z} \cdot y^{z - x} \cdot z^{x - y} = 1: Sequence & Series (Mathematics)

$\begin{aligned} &\text { Then } x=a+(m-1) d \text { and } x=b r^{m-1}\\ & \quad y=a+(n-1) d \text { and } y=b r^{n-1} \\& \quad z=a+(p-1) d \text { and } z=b r^{p-1} \\ & \therefore x-y=(m-n) d, y-z=(n-p) d, z-x=(p-m) d \\& \text { Now } x^{y-z} y^{z-x} z^{x-y}=\left[b r^{m-1}\right]^{(n-p) d}\left[b r^{n-1}\right]^{(p-m) d}\left[b r^{p-1}\right]^{(m-n) d}= \\ & b^{[n-p+p-m+m-n] d} r^{[(m-1)(n-p)+(n-1)(p-m)+(p-1)(m-n)] d}=b^{0 \cdot d} r^{0 \cdot d}=1\end{aligned}$

Explanation

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

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