If _32,_3( 2^x - 5 ) and _3( 2^x - 72 ) are in arithmetic progression, determine the value of x

2 _3(2^x-5)= _3 2+ _3(2^x-72) \\ (2^x-5)^2=2 (2^x-72) \\ (2^x)^2-10 2^x+25=2 2^x-7 is a quadratic equation in 2^x . \\ (2^x)^2-12 2^x+32=0 \\ (2^x-4)(2^x-8)=0 \\

If \log_{3}2,\log_{3}( 2^{x} - 5 ) and \log_{3}( 2^{x} - \frac{7}{2} ) are in arithmetic progression, determine the value of x: Sequence & Series (Mathematics)

$2 \log _3\left(2^x-5\right)=\log _3 2+\log _3\left(2^x-\frac{7}{2}\right)$ $\\ \therefore\left(2^x-5\right)^2=2 \times\left(2^x-\frac{7}{2}\right) \\$ $\therefore\left(2^x\right)^2-10 \times 2^x+25=2 \times 2^x-7$ { is a quadratic equation in } $2^x . \\$ $\therefore\left(2^x\right)^2-12 \times 2^x+32=0 \\$ $\therefore\left(2^x-4\right)\left(2^x-8\right)=0 \\$ $\therefore 2^x=4$ { or } $2^x=8 \therefore x=2$ { or } 3 For $x=2, \log _3\left(2^x-5\right)\\$ is not meaningful $\therefore x=3$ is the real solution.

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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