1. **Problem Statement:** We are given the 8th term ($T_8$) and the 17th term ($T_{17}$) of a Harmonic Progression (HP) as $\frac{2}{5}$ and $\frac{2}{11}$ respectively. We need to find the 35th term ($T_{35}$) of the HP. 2. **Recall the definition and formula:** In a Harmonic Progression, the reciprocals of the terms form an Arithmetic Progression (AP). If $a_n$ is the $n^{th}$ term of the HP, then $\frac{1}{a_n}$ forms an AP. 3. **Set up the AP for reciprocals:** Let $b_n = \frac{1}{a_n}$ be the $n^{th}$ term of the AP. Then, $$b_8 = \frac{1}{T_8} = \frac{1}{\frac{2}{5}} = \frac{5}{2}$$ $$b_{17} = \frac{1}{T_{17}} = \frac{1}{\frac{2}{11}} = \frac{11}{2}$$ 4. **Use the AP formula:** The $n^{th}$ term of an AP is $$b_n = b_1 + (n-1)d$$ where $b_1$ is the first term and $d$ is the common difference. 5. **Write equations for $b_8$ and $b_{17}$:** $$b_8 = b_1 + 7d = \frac{5}{2}$$ $$b_{17} = b_1 + 16d = \frac{11}{2}$$ 6. **Subtract the first from the second to find $d$:** $$b_{17} - b_8 = (b_1 + 16d) - (b_1 + 7d) = 9d = \frac{11}{2} - \frac{5}{2} = \frac{6}{2} = 3$$ $$\Rightarrow d = \frac{3}{9} = \frac{1}{3}$$ 7. **Find $b_1$ using $b_8$:** $$b_1 + 7 \times \frac{1}{3} = \frac{5}{2}$$ $$b_1 + \frac{7}{3} = \frac{5}{2}$$ $$b_1 = \frac{5}{2} - \frac{7}{3} = \frac{15}{6} - \frac{14}{6} = \frac{1}{6}$$ 8. **Find $b_{35}$:** $$b_{35} = b_1 + 34d = \frac{1}{6} + 34 \times \frac{1}{3} = \frac{1}{6} + \frac{34}{3} = \frac{1}{6} + \frac{68}{6} = \frac{69}{6} = \frac{23}{2}$$ 9. **Find $T_{35}$:** Since $T_{35} = \frac{1}{b_{35}}$, $$T_{35} = \frac{1}{\frac{23}{2}} = \frac{2}{23}$$ **Final answer:** The 35th term of the HP is $\boxed{\frac{2}{23}}$.