The common difference of the sequence is

$\displaystyle 3 \frac{1}{4} - 1 \frac{1}{7} = 2\frac{3}{28}$

so the $\displaystyle n$th term of the sequence is

$\displaystyle a_{n} = 1\frac{1}{7} + \left ( 2\frac{3}{28} \right ) (n-1)$

To find out the minimum value for which $\displaystyle a_{n} > 100$, set up this inequality:

$\displaystyle 1\frac{1}{7} + \left ( 2\frac{3}{28} \right ) (n-1) > 100$

$\displaystyle 1\frac{1}{7} + \left ( 2\frac{3}{28} \right ) n -2\frac{3}{28} > 100$

$\displaystyle \left ( 2\frac{3}{28} \right ) n - \frac{27}{28} > 100$

$\displaystyle \left ( 2\frac{3}{28} \right ) n > 100 \frac{27}{28}$

$\displaystyle n > 100 \frac{27}{28} \div 2\frac{3}{28} = 47 \frac{54}{59}$

The correct response is the forty-eighth term.

The $\displaystyle n$th term of an arithmetic sequence with initial term $\displaystyle a_{1}$ and common difference $\displaystyle d$ is defined by the equation

$\displaystyle a_{n} = a_{1}+ (n-1)d$

Since the tenth and twelfth terms are two terms apart, the common difference can be found as follows:

$\displaystyle a_{10}+2d=a_{12}$

$\displaystyle 8.4+2d=10.2$

$\displaystyle 8.4+2d- 8.4=10.2 - 8.4$

$\displaystyle 2d=1.8$

$\displaystyle d = 0.9$

Now, we can set $\displaystyle n = 10, d = 0.9$ in the sequence equation to find $\displaystyle a_{1}$:

$\displaystyle a_{n} = a_{1}+ (n-1)d$

$\displaystyle a_{10} = a_{1}+ (10-1)0.9$

$\displaystyle 8.4 = a_{1}+ 9 \cdot 0.9$

$\displaystyle 8.4 = a_{1}+ 8.1$

$\displaystyle 8.4 - 8.1 = a_{1}+ 8.1 - 8.1$

$\displaystyle a_{1} = 0.3$

The $\displaystyle n$th term of an arithmetic sequence with initial term $\displaystyle a_{1}$ and common difference $\displaystyle d$ is defined by the equation

$\displaystyle a_{n} = a_{1}+ (n-1)d$

Since the eleventh and thirteenth terms are two terms apart, the common difference can be found as follows:

$\displaystyle a_{11}+2d=a_{13}$

$\displaystyle 11+2d=14$

$\displaystyle 11+2d -11=14 -11$

$\displaystyle 2d = 3$

$\displaystyle d = \frac{3}{2}$

Now, we can set $\displaystyle n = 11, d =\frac{3}{2}$ in the sequence equation to find $\displaystyle a_{1}$:

$\displaystyle a_{11} = a_{1}+ (11-1) \frac{3}{2}$

$\displaystyle 11= a_{1}+ 10 \cdot \frac{3}{2}$

$\displaystyle 11= a_{1}+ 15$

$\displaystyle 11- 15= a_{1}+ 15 - 15$

$\displaystyle a_{1} = -4$

Let $\displaystyle a_{n}$ be the lengths of the sides of the squares in inches. $\displaystyle a_{1} = 8$ and $\displaystyle a_{2} = 12$, so their common difference is

$\displaystyle d = a_{2} - a_{1} = 12 - 8 = 4$

The arithmetic sequence formula is 

$\displaystyle a_{n} = a_{1} + (n-1)d$

The length of a side of the largest square - square 10 - can be found by substituting $\displaystyle a_{1} = 8, n= 10, d = 4$:

$\displaystyle a_{10} =8+ (10-1) 4 = 8 + 9 \cdot 4 = 8 + 36 = 44$ 

The largest square has sides of length 44 inches, so its area is the square of this, or $\displaystyle A = 44^{2} = 1,936$ square inches.

The $\displaystyle n$th term of an arithmetic sequence with initial term $\displaystyle a_{1}$ and common difference $\displaystyle d$ is defined by the equation

$\displaystyle a_{n} = a_{1}+ (n-1)d$

The initial term in the given sequence is

$\displaystyle a_{1} = 12$;

the common difference is

$\displaystyle d = 19-12= 7$;

We are seeking term $\displaystyle n = 32$.

This term is  

$\displaystyle a_{32} = 12+ (32-1) 7$

$\displaystyle a_{32} = 12+ 31 \cdot 7 = 12+ 217 = 229$

The $\displaystyle n$th term of an arithmetic sequence with initial term $\displaystyle a_{1}$ and common difference $\displaystyle d$ is defined by the equation

$\displaystyle a_{n} = a_{1}+ (n-1)d$.

The initial term in the given sequence is

$\displaystyle a_{1} = \frac{2}{5}$;

the common difference is

$\displaystyle d = \frac{7}{10} - \frac{2}{5} = \frac{7}{10} - \frac{4}{10} = \frac{3}{10}$.

We are seeking term $\displaystyle n = 33$.

Therefore,

$\displaystyle a_{33} = \frac{2}{5}+ (33-1) \frac{3}{10}$

$\displaystyle = \frac{2}{5}+ (32) \frac{3}{10}$

$\displaystyle = \frac{2}{5}+ \frac{96}{10}$

$\displaystyle = \frac{2}{5}+ \frac{48}{5}$

$\displaystyle = \frac{50}{5} = 10$,

which is not among the choices.

In this sequence, each subsequent number is equal to one third of the preceding number. 

One third of 11 is equal to:

$\displaystyle \frac{11}{3}=3\frac{2}{3}$

Therefore, the correct answer is: $\displaystyle 3\frac{2}{3}$

The common difference for this sequence is $\displaystyle 5$. To find the next number in the sequence, add $\displaystyle 5$ to the last given number.

$\displaystyle 34+5=39$

The common difference for this sequence is $\displaystyle 7$. Add this to the last given term to find the next one.

$\displaystyle -1+7=6$

The common difference is $\displaystyle 9$. Add this to the last given term to find the next term.

$\displaystyle 14+9=23$