The sum of the first three terms of a geometric sequence is equal to 42. The sum of the squares of the same terms is equal to 1092. Can you Find the three terms of the sequence?

Answer 1

The solution to this problem is {32, 8, 2}. I solved it by graphing two functions derived from the problem statement, and then finding out where the two functions intersect.

Answer 2

This is the sum of a geometric series problem of the form a1 + a1 r + a1 r² + a1 r³

Note if a1 = 2 and r = 4 We have 2 + 8 + 32 = 42 Then if we cube this expression we have:

4 + 64 + 1024 = 1092

Answer 3

This is the sum of a geometric series problem of the form a1 + a1 r + a1 r² + a1 r³

Note if a1 = 2 and r = 4 We have 2 + 8 + 32 = 42 Then if we cube this expression we have:

4 + 64 + 1024 = 1092