What is the sum of a geometric sequence 1, 3, 9, if there are 12 terms? a. 292,524b. 265,720c. 139,968d. 104,976

We know that
a1=1
a2=3
a3=9
a2/a1=3/1> 3
a3/a2=9/3> 3
common ration r is equal to 3
number of terms n  is 12

The Sum of geometric series is given by the formula
Sum=a1*[1-r^n]/[1-r]
Sum=1*[1-3^12]/[1-3]> Sum=[1-3^12]/[1-3]> [3^12-1]/[3-1]
Sum=531440/2> 265720

the answer is
265720

The first one, i believe, would be correct. all the others have one thing in common, and that's that they have repeating numbers. in the second one, x and y have repeating 6's and 3's, and in the third, x and y have repeating 6's as well. in the last one, there are repeating 3's. the first one has no repeating numbers.