What is the common difference of an arithmetic sequence if the sum of the second and fourth terms is 0, and the sum of the sixth and eighth terms is 12?
3 answers
I'm really confused on this problem. If the sum of the second and fourth terms is 0 then the two terms cannot be 0 but a negative number that crosses each other out
first condition:
a+d + a+3d = 0
2a + 4d = 0
a + 2d = 0
a = -2d
a + 5d + a+ 7d = 12
2a + 12d = 12
a + 6d = 6
sub in the first part
-2d + 6d = 6
4d = 6
d = 3/2
then a = -2(3/2) = -3
check:
2nd term = -3+3/2 = -3/2
4th term = -3 + 3(3/2) = 3/2
their sum = -3/2 + 3/2 = 0
6th term = -3 + 5(3/2) = 9/2
8th term = -3 + 7(3/2) = 15/2
their sum = 9/2 + 15/2 = 12
a+d + a+3d = 0
2a + 4d = 0
a + 2d = 0
a = -2d
a + 5d + a+ 7d = 12
2a + 12d = 12
a + 6d = 6
sub in the first part
-2d + 6d = 6
4d = 6
d = 3/2
then a = -2(3/2) = -3
check:
2nd term = -3+3/2 = -3/2
4th term = -3 + 3(3/2) = 3/2
their sum = -3/2 + 3/2 = 0
6th term = -3 + 5(3/2) = 9/2
8th term = -3 + 7(3/2) = 15/2
their sum = 9/2 + 15/2 = 12
Thanks a lot.
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