If a, b, and c are in arithmetic progression, | vedclass

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(D) Given that $a, b,$ and $c$ are in arithmetic progression $(AP)$.Therefore,$2b = a + c$.Consider the terms $2^{ax + 1}, 2^{bx + 1},$ and $2^{cx + 1

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in geometric progression for every $x 
eq 0$.

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(D) Given that $a, b,$ and $c$ are in arithmetic progression $(AP)$.Therefore,$2b = a + c$.Consider the terms $2^{ax + 1}, 2^{bx + 1},$ and $2^{cx + 1}$.For these terms to be in geometric progression $(GP)$,the ratio of consecutive terms must be equal:$\frac{2^{bx + 1}}{2^{ax + 1}} = \frac{2^{cx + 1}}{2^{bx + 1}}$$2^{(b - a)x} = 2^{(c - b)x}$Since $a, b, c$ are in $AP$,we have $b - a = c - b = d$ (common difference).Thus,$2^{dx} = 2^{dx}$,which is always true for any $x 
eq 0$.Therefore,$2^{ax + 1}, 2^{bx + 1},$ and $2^{cx + 1}$ are in geometric progression for every $x 
eq 0$.

If $a, b,$ and $c$ are in arithmetic progression,then $2^{ax + 1}, 2^{bx + 1},$ and $2^{cx + 1}$ for $x \neq 0$ are...

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