Chapter 11.3, Problem 9P

a.

To determine

To calculate: The Numbers of the bricks are in the 12^th row.

Answer to Problem 9P

The numbers of bricks are in 12^th row is 19.

Explanation of Solution

Given information: The numbers of bricks are in last row = 85

The numbers of bricks are in second last row = 79

The numbers of bricks are in third last row = 73

Since, $85,79,73,\dots \dots \dots \mathrm{..}\dots 1$

Hence, the above series are in A.P

The first term, $a=85$

Second terms, $a_1=79$

The common difference, $d=a_1-a=79-85=-6$

Formula used: $a_n=a+\left(n-1\right)d$, where n is numbers of terms

Now, 12^th term

Calculation:

  $\begin{array}{l}\therefore a_{12}=85+\left(12-1\right)\left(-6\right)\\ =85+\left(11\right)\times \left(-6\right)\\ =85-66\\ =19\end{array}$

Hence, the 12^th term is 19.

b.

To determine

To calculate: The total numbers of rows.

Answer to Problem 9P

Total numbers of rows are $15$ .

Explanation of Solution

Given information: The numbers of bricks are in last row = 85

The numbers of bricks are in second last row = 79

The numbers of bricks are in third last row = 73

The numbers of bricks are in top most = 1

Since, $85,79,73,\dots \dots \dots \mathrm{..}\dots 1$

Hence, the above series are in A.P

The first term, $a=85$

Second terms, $a_1=79$

Last term, $l=1$

The common difference, $d=a_1-a=79-85=-6$

Formula used: $l=a+\left(n-1\right)d$, where n is numbers of rows

Calculation:

  $\begin{array}{l}\therefore 1=85+\left(n-1\right)\left(-6\right)\\ \Rightarrow 1-85=-6n+6\\ \Rightarrow 6n=84+6\\ \Rightarrow n=\frac{90}{6}=15\\ \therefore n=6\end{array}$

Hence, the total numbers of rows are $15$ .