Chapter 7.9, Problem 12CYU

To determine

To Find: The $nth$ term and ninth term of geometric sequence $112,84,\mathrm{63............}$ .

Answer to Problem 12CYU

  $a_n=\left(112\right){\left(\frac{3}{4}\right)}^{n-1}\&a_9=\frac{45927}{4096}$

Explanation of Solution

Given information:

Geometric sequence is $112,84,\mathrm{63............}$ .

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 $112,84,\mathrm{63............}$

Common ratio is

  $\begin{array}{l}r=\frac{84}{112}=\frac{63}{84}=\frac{3}{4},a=112\\ \text{Then }nth\text{ term of the sequence is}\\ a_n=a{r}^{n-1}\\ a_n=\left(112\right){\left(\frac{3}{4}\right)}^{n-1}\\ a_n=112{\left(\frac{3}{4}\right)}^{n-1}\end{array}$

Now, the ninth term is

  $\begin{array}{l}a=112,r=\frac{3}{4}\&n=9\\ a_9=112{\left(\frac{3}{4}\right)}^{9-1}\\ a_9=112{\left(\frac{3}{4}\right)}^8\\ a_9=7\times \frac{3^8}{4^6}\\ a_9=\frac{45927}{4096}\end{array}$

Conclusion:

Therefore, the required terms for the given series are $a_n=\left(112\right){\left(\frac{3}{4}\right)}^{n-1}\&a_9=\frac{45927}{4096}$.