A group of women working together at the same rate can build a wall...

Let $n$ women complete the work in 45 hr . Then 1 woman can complete the work in $45 n \mathrm{hr}$. $\therefore$ Each woman's work of $1 \mathrm{hr}=\frac{1}{45 n}$. According to given question let each woman join the work after $d$ hrs. of previous women started the work. Now, total work is finished in $T$ hours after the work started by first woman. Time for which the first woman work $=T \mathrm{hr}$. Time for which the $n^{\text {th }}$ women work $=\{T-(n-1) d\} \mathrm{hr}$. A.T.Q : $5\{T-(n-1) d\}=T$ $$\begin{aligned} & \therefore \frac{4 T}{5}=(n-1) d \\ & \text { and } \frac{T}{45 n}+\frac{T-d}{45 n}+\frac{T-2 d}{45 n}+\ldots+\frac{T-(n-1) d}{45 n}=1 \\ & \Rightarrow n T-\frac{n(n-1)}{2} d=45 n \\ & \Rightarrow 2 T-(n-1) d=90 \\ & \Rightarrow 2 T-\frac{4 T}{5}=90 \\ & \Rightarrow \frac{6 T}{5}=90 \\ & \therefore T=\frac{90 \times 5}{6}=75 \end{aligned}$$ $\therefore$ Total time for which first women work $=75 \mathrm{hrs}$