Question

What is the 10th term of the following geometric sequence? -7/9, 7/3, -7, 21, -63

Answer 1

The geometric sequence has a formula that is $a_{n} = a_{1} (r) ^{n-1}$, where n is the number of the term in question, a1 is the first number in the sequence, and r is the common ratio.  If we inspect those numbers we have thus far, we see that they alternate between negative and positive and negative, etc.  Therefore, the common ratio has to be a negative number.  In order to get from -7/9 to +7/3, we would have to multiply by -3: $- \frac{7}{9} *- \frac{3}{1} = \frac{7}{3}$ just as proof.  r then is -3.  We are looking for n = 10, so our formula is $a_{10} =- \frac{7}{9}(-3) ^{10-1}$.  Simplifying a bit we have $a_{10}=- \frac{7}{9}(-3)^9$ and $a_{10} =- \frac{7}{9}(-19683)$.  That means that the tenth term in that sequence is 2187