If $7=5+\frac{1}{7}(5+\alpha )+\frac{1}{{7}^{2}}(5+2\alpha )$$+\frac{1}{{7}^{3}}(5+3\alpha )+\ldots \ldots \ldots ...\infty$ then the value of $\alpha$ is : [JEE Main 2025, 24 Jan (Shift 2)]

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If \(7=5+\frac{1}{7}(5+\alpha )+\frac{1}{{7}^{2}}(5+2\alpha )\)\(+\frac{1}{{7}^{3}}(5+3\alpha )+\ldots \ldots \ldots ...…

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