【Solution】Solution: Let the later portions be: $a, 2a, 3a, 4a, 5a, 6a$. Then their original values were $a+150, 2a-10, 3a-20, 4a-30, 5a-40, 6a-50$. According to the subsequent numbers, the common difference is $5a-40-(4a-30)=a-10$. Based on the common difference, the value before $2a-10$ should be $a-20$. Therefore, $a+150$ is the maximum value of the sequence. $a+150-(a-10)=160$. So, $6a-50=160$. Therefore, $a=35$. Hence, the later quantities are $35, 70, 105, 140, 175, 210$. The total is $35+70+105+140+175+210=735$ (units). Therefore, the answer is: 735
9. (16 points) There are six piles of apples, and their numbers form an arithmetic sequence. Junjun picks one of the piles and takes out 150 apples, distributing them to the other 5 piles, giving 10, 20, 30, 40, and 50 apples to each pile in sequence. After the distribution, Junjun finds that the number of apples in these 5 piles is exactly 2 times, 3 times, 4 times, 5 times, and 6 times the number of apples in the pile he picked. How many apples are there in total in the six piles? $\qquad$