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 * y^z - x * z^x - y = 1

Then x=a+(m-1) d and x=b r^m-1\ y=a+(n-1) d and y=b r^n-1 z=a+(p-1) d and z=b r^p-1 \ Therefore x-y=(m-n) d, y-z=(n-p) d, z-x=(p-m) d Now x^y-z y^z-x z^x-y=[b r^m-1]^(n-p) d[b r^n-1]^(p-m) d[b r^p-1]^(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 * d r^0 * d=1\

If the m^{{th}{~}},n^{{th}{~}} and p^{{th}{~}} terms ... | 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

Let a_{1},a_{2},a_{3},. be a G.P. of increasing positive numbers. Let the sum of its 6^{{th}{~}} and 8^{{th}{~}} terms be 2 and the product of its 3^{{rd}{~}} and 5^{{th}{~}} terms be 1/9. Then 6( a_{2} + a_{4} )( a_{4} + a_{6} ) is equal to

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Let \frac{1}{x_{1}},\frac{1}{x_{2}},,\frac{1}{x_{n}}( x_{i} eq 0 .\ for . \ i = 1,2,,n ) be in A.P. such that x_{1} = 4 and x_{21} = 20. If n is the least positive integer for which x_{n} > 50, then \sum_{i = 1}^{n}\mspace{2mu}( \frac{1}{x_{i}} ) is equal to

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