You are given $e_{60}=17$, $_{10}P_50=0.8$, and that the $10$-year curtate temporary life expectancy at age $50$ is $9.2$. Find $e_{50}$.
$e_x=P_{x}(1+e_{x+1})$
So,
$e_{59}=P_{59}(1+e_{60}) \rightarrow e_{59}=18(P_{59}) $
$e_{58}=P_{58}(1+e_{59})=P_{58}(1+_{18}P_{59})=P_{58}+18P_{58}.P_{59}=_1P_{58}+18_2P_{58}$
$e_{57}=P_{57}(1+e_{58})=P_{57}(1+P_{58}+18_2P_{58})=P_{57}+_2P_{57}+18_2P_{58}=_1P_{57}+_2P_{57}+18._3P_{57}$
Similarly, by recursion,
$e_{50}=\sum_{k=1}^9(_kP_x)+18(_{10}P_{50}) \rightarrow e_{50}=23.6 $
But the answer is $22.8$
$\require{enclose}$ We have: $$e_{60} = 17, \quad {}_{10} p_{50} = 0.8, \quad e_{50 : \enclose{actuarial}{10}} = 9.2,$$ and we wish to determine $e_{50}$. We know that $$e_x = \sum_{k=1}^\infty {}_k p_x, \quad e_{x:\enclose{actuarial}{n}} = \sum_{k=1}^n {}_k p_x.$$ So it is natural to conclude that $$\begin{align*} e_{50} &= \sum_{k=1}^{10} {}_k p_{50} + \sum_{k=11}^\infty {}_k p_{50} \\ &= e_{50:\enclose{actuarial}{10}} + \sum_{k=1}^\infty {}_k p_{60} \,_{10} p_{50} \\ &= e_{50:\enclose{actuarial}{10}} + {}_{10} p_{50} \, e_{60} \\ &= 9.2 + (0.8)(17) \\ &= 22.8. \end{align*} $$
It is worth mentioning that you should not use ${}_k P_x$ when you mean ${}_k p_x$: the latter is the $k$-year survival probability of a life aged $x$, but the former represents a fully discrete annual benefit premium of a $k$-payment whole life insurance, i.e., $${}_k P_x = \frac{A_x}{\ddot a_{x : \enclose{actuarial}{k}}}.$$ Actuarial notation is case-sensitive.