+26376 Find the 10th term of the sequence 2, 8, 32, 128....
The general form of a geometric sequence is
\( {\displaystyle a,\ ar,\ ar^{2},\ ar^{3},\ ar^{4},\ \ldots } \)
The common ratio r is \(\frac{8}{2}=\frac{32}{8}=\frac{128}{32}=4\)
a is a scale factor, equal to the sequence's start value = 2
The n-th term of a geometric sequence with initial value a and common ratio r is given by
\(\begin{array}{rcll} a_{n} &=& a\,r^{n-1} \quad &| \quad a=2 \quad r = 4 \\ a_{n} &=& 2\cdot 4^{n-1} \\ \end{array}\)
The 10th term is
\(\begin{array}{|rcll|} \hline a_{10} &=& 2\cdot 4^{10-1} \\ a_{10} &=& 2\cdot 4^{9} \\ a_{10} &=& 2\cdot 262144 \\ \mathbf{ a_{10}} & \mathbf{=} & \mathbf{524288} \\ \hline \end{array}\)
+128731 I think this should be 2, 8, 32 , 128.....
The 10h term is given by : 22n - 1 = 22(10) - 1 = 219 = 524,298
+26376 Find the 10th term of the sequence 2, 8, 32, 128....
The general form of a geometric sequence is
\( {\displaystyle a,\ ar,\ ar^{2},\ ar^{3},\ ar^{4},\ \ldots } \)
The common ratio r is \(\frac{8}{2}=\frac{32}{8}=\frac{128}{32}=4\)
a is a scale factor, equal to the sequence's start value = 2
The n-th term of a geometric sequence with initial value a and common ratio r is given by
\(\begin{array}{rcll} a_{n} &=& a\,r^{n-1} \quad &| \quad a=2 \quad r = 4 \\ a_{n} &=& 2\cdot 4^{n-1} \\ \end{array}\)
The 10th term is
\(\begin{array}{|rcll|} \hline a_{10} &=& 2\cdot 4^{10-1} \\ a_{10} &=& 2\cdot 4^{9} \\ a_{10} &=& 2\cdot 262144 \\ \mathbf{ a_{10}} & \mathbf{=} & \mathbf{524288} \\ \hline \end{array}\)
