If the sum of the series$\text{ }\frac{1}{1\cdot (1+d)}+\frac{1}{(1+d)(1+2d)}+\ldots +\frac{1}{(1+9d)(1+10d)}\text{ , }$is equal to $5$ then $50d$ is equal to : [JEE Main 2024, 9 Apr (Shift 1)]

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If the sum of the series\(\text{ }\frac{1}{1\cdot (1+d)}+\frac{1}{(1+d)(1+2d)}+\ldots +\frac{1}{(1+9d)(1+10d)}\text{ , }…

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