What is the first term of the geometric sequence presented in the table below?
n 3 6
an 12 −324
Hint: an = a1(r)n − 1, where a1 is the first term and r is the common ratio.
+129845 12 = a1(r)2 → a1 = 12/r2 (1)
-324 = a1(r)5 (2)
Putting (1) into (2), we have
-324 = (12/r2)(r)5
-324/12 = r3
-27 = r3
r = (-27)1/3 = -3
So
a1 = 12/(-3)2 = 12/9 = 4/3
+26387 What is the first term of the geometric sequence presented in the table below?
n 3 6
an 12 −324
\(\small{ \boxed{~ \text{geometric sequence: } \quad a_k = a_i^{ \frac{j-k}{j-i} }\cdot a_j^{ \frac{k-i}{j-i} } ~} } \)
\(\begin{array}{rcll} a_3 &=& a_i = 12 \qquad & i = 3\\ a_6 &=& a_j =-324 \qquad & j = 6\\ a_1 &=& a_k = \ ? \qquad & k = 1 \end{array}\)
\(\begin{array}{rcll} a_1 &=& 12 ^{ \left( \frac{6-1}{6-3} \right)} \cdot (-324)^{ \left( \frac{1-3}{6-3} \right) } \\ a_1 &=& 12 ^{\left( \frac{5}{3} \right) } \cdot (-324)^{ \left( \frac{-2}{3} \right) } \\ a_1 &=& \frac{ 12 ^{\left( \frac{5}{3} \right) } } { (-324)^{ \left( \frac{2}{3} \right) } }\\ a_1 &=& \sqrt[3]{ \frac{ 12^5 } { (-324)^{2} } } \\ a_1 &=& \sqrt[3]{ \frac{ 248832 } { 104976 } } \\ a_1 &=& \sqrt[3]{ 2.37037037037 } \\ a_1 &=& 1.33333333333 \\ a_1 &=& \frac43 \\ \end{array}\)
