Chapter 7.9, Problem 11CYU

To determine

To Find: The $nth$ term and tenth term of geometric sequence $72,48,\mathrm{32.............}$ .

Answer to Problem 11CYU

  $a_n=72{\left(\frac{2}{3}\right)}^{n-1}\&a_{10}=\frac{4096}{2187}$

Explanation of Solution

Given information:

Geometric sequence is $72,48,\mathrm{32.............}$ .

Concept and Formula Used:

Geometric Sequence: Sequence in which each term after the first is found by multiplying the previous term by a constant called common ratio.

  $nth$ term of the geometric sequence is $a_n=a{r}^{n-1}$ where ‘a’ is the first term and ‘r’ is the common ratio.

Calculation:

It is given that geometric sequence is $72,48,\mathrm{32.............}$

Common ratio is

  $\begin{array}{l}r=\frac{48}{72}=\frac{32}{48}=\frac{2}{3},a=72\\ \text{Then }nth\text{ term of the sequence is}\\ a_n=\left(72\right){\left(\frac{2}{3}\right)}^{n-1}\\ a_n=72{\left(\frac{2}{3}\right)}^{n-1}\end{array}$

Now, the tenth term is

  $\begin{array}{l}a=-1,r=-5\&n=7\\ a_{10}=72{\left(\frac{2}{3}\right)}^{10-1}\\ a_{10}=72{\left(\frac{2}{3}\right)}^9\\ a_{10}=72\times \frac{2^9}{3^9}\\ a_{10}=\frac{4096}{2187}\end{array}$

Conclusion:

  $a_n=72{\left(\frac{2}{3}\right)}^{n-1}\&a_{10}=\frac{4096}{2187}$