# Finding the ${n}^{\text{th}}$ Term of a Geometric Sequence

Given a geometric sequence with the first term ${a}_{1}$ and the common ratio $r$ , the ${n}^{\text{th}}$ (or general) term is given by
${a}_{n}={a}_{1}\cdot {r}^{n-1}$ .

Example 1:

Find the ${6}^{\text{th}}$ term in the geometric sequence $3,12,48,...$ .

$\begin{array}{l}{a}_{1}=3,\text{\hspace{0.17em}}\text{\hspace{0.17em}}r=\frac{12}{3}=4\\ {a}_{6}=3\cdot {4}^{6-1}=3\cdot {4}^{5}=3072\end{array}$

Example 2:

Find the ${7}^{\text{th}}$ term for the geometric sequence in which ${a}_{2}=24$ and ${a}_{5}=3$ .

Substitute $24$ for ${a}_{2}$ and $3$ for ${a}_{5}$ in the formula

${a}_{n}={a}_{1}\cdot {r}^{n-1}$ .

$\begin{array}{l}{a}_{2}={a}_{1}\cdot {r}^{2-1}\to 24={a}_{1}r\\ {a}_{5}={a}_{1}\cdot {r}^{5-1}\to \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3={a}_{1}{r}^{4}\end{array}$

Solve the firstequation for ${a}_{1}$ : ${a}_{1}=\frac{24}{r}$

Substitute this expression for ${a}_{1}$ in the second equation and solve for $r$ .

Substitute for $r$ in the first equation and solve for ${a}_{1}$ .

$\begin{array}{l}24={a}_{1}\left(\frac{1}{2}\right)\\ 48={a}_{1}\end{array}$

Now use the formula to find ${a}_{7}$ .

${a}_{7}=48{\left(\frac{1}{2}\right)}^{7-1}=48\cdot \frac{1}{64}=\frac{3}{4}$