Question

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?

Answer 1

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 + 2b²) / 3 = 394
(150+b²)=591
b²=441
b=-21,21

thus the numbers will be:
10-21, 10, 10+21
=-11,10,31