Chapter 9.4, Problem 8QR

Solution

To calculate: The value of $a_{10}$ if $a_8=10;a_{12}=40$ .

The resultant value of the tenth term is 20.

Given information:

The eighth term is 10 and twelfth term is 40.

Formula used: The formula for the nth term in a geometric sequence is:

  $a_n=a_1\cdot r^{n-1}$

Calculation:

Consider the given information,

  $a_8=10;a_{12}=40$

The eighth term can be written as:

  $\begin{array}{l}{a}_8=a_1\cdot r^{8-1}\\ a_8=a_1\cdot r^7\end{array}$

The twelfth term can be written as:

  $\begin{array}{l}{a}_{12}=a_1\cdot r^{12-1}\\ a_{12}=a_1\cdot r^{11}\end{array}$

Now find the common ratio ( r ):

  $\begin{array}{c}\frac{a_{12}}{a_8}=\frac{a_1\cdot r^{11}}{a_1\cdot r^7}\\ \frac{40}{10}=\frac{a_1\cdot r^{11}}{a_1\cdot r^7}\\ 4=r^4\\ \sqrt{2}=r\end{array}$

Find the value of the first term:

  $\begin{array}{l}{a}_8=a_1\cdot r^7\\ 10=a_1\cdot {\left(\sqrt{2}\right)}^7\\ 10=a_1\cdot \left(8\sqrt{2}\right)\\ \frac{10}{8\sqrt{2}}=a_1\end{array}$

To find the value of $a_{10}$ substitute the values $a_1=\frac{10}{8\sqrt{2}},r=\sqrt{2},n=10$ in $a_n=a_1\cdot r^{n-1}$ ,

  $\begin{array}{c}{a}_n=a_1\cdot r^{n-1}\\ a_{10}=\frac{10}{8\sqrt{2}}\cdot {\left(\sqrt{2}\right)}^{10-1}\\ a_{10}=\frac{10}{8\sqrt{2}}\cdot {\left(\sqrt{2}\right)}^9\\ a_{10}=\frac{10}{8\sqrt{2}}\cdot 16\sqrt{2}\\ a_{10}=20\end{array}$

Thus, the value of $a_{10}$ is 20.