The mean of 3 consecutive terms in an arithmetic sequence is 10, and the mean of their squares is 394. What is the largest of the 3 terms?

Let the numbers be:

a, a+b, a+2b

mean of the numbers is:

(a+a+b+a+2b) / 3=10

(3a+3b) / 3=10

a+b=10

thus second term is 10

The sum of square will be:

((10 - b) ² + 10² + (10 + b) ²) / 3 = 394

(100 - 20b + b² + 100 + 100 + 20b + b²) / 3 = 394

(300 + 2b ²) / 3 = 394

(150+b²) = 591

b²=441

b=-21,21

thus the numbers will be:

10-21, 10, 10+21

=-11,10,31