An arithmetic sequence is one in which the difference of each term in the sequence and the one preceding is constant.

The two statements together demonstrate that two such differences are equal to each other, and that two other such differences are equal to each other. No information, however, is given about any other differences, of which there are infinitely many. Therefore, the question of whether the sequence is arithmetic or not is unresolved.

Each term of an arithmetic sequence is the preceding term plus the same number, the common difference.

Assume Statement 1 alone, and examine the sequences:

$\displaystyle 12, 20, 28,...$

$\displaystyle 9, 19, 29,...$

Both sequences are arithmetic; the first has common difference 8, the second, common difference 9. In both sequences, the arithmetic mean of the second and third terms - half their sum - is $\displaystyle \frac{20+28}{2} = \frac{19+29}{2} = 24$. However, the first term differs.

Assume Statement 2 alone. The common difference alone is not enough to determine the first term, as evidenced in these two sequences:

$\displaystyle 10, 20, 30,...$

$\displaystyle 11,21, 31, ...$

both of which have common difference 10.

Now assume both statements. The arithmetic mean of $\displaystyle a_{2}$ and $\displaystyle a_{3}$ is 24, so 

$\displaystyle \frac{a_{2}+a_{3}}{2} = 24$

or 

$\displaystyle \frac{1}{2}a_{2}+ \frac{1}{2}a_{3} = 24$

Also, the common difference is 10, so

$\displaystyle a_{3}-a_{2} = 10$

These two equations form a two-by-two linear system which can be solved as follows:

$\displaystyle \left (\frac{1}{2}a_{2}+ \frac{1}{2}a_{3} \right )\cdot 2 = 24 \cdot 2$

$\displaystyle a_{3}+a_{2} = 48$

$\displaystyle \underline{a_{3}-a_{2} = 10}$

$\displaystyle 2a_{3} \; \; \; \; \; \; = 58$

$\displaystyle 2a_{3} \div 2 = 58 \div 2$

$\displaystyle a_{3}= 29$

$\displaystyle a_{2} = a_{3} - 10 = 29 - 10 = 19$

$\displaystyle a_{1} = a_{2} - 10 = 19 - 10 = 9$

$\displaystyle g(x) = f(x- 3)$ in each statement, so the graph of the function $\displaystyle g$ is the same as that of the function $\displaystyle f$ translated three units right. However, we are restricting the domain of $\displaystyle g$ to $\displaystyle [-3, 3]$. Each of the two statements examines one half of the graph. See the graph below, which divides the graph into the portion on the domain $\displaystyle [-3, 0]$ (in blue; discussed in Statement 1) and the portion on the domain $\displaystyle [0, 3]$ (in green, discussed in Statement 2):

Assume Statement 1 alone. The portion of the graph of $\displaystyle g$ on the domain $\displaystyle [-3, 0]$ passes the horizontal line test, since no horizontal line passes through it twice. However, without knowing anything about the other half of the graph, the question about whether $\displaystyle g^{-1}$ exists cannot be resolved. 

Assume Statement 2 alone. Notice that we can draw a horizontal line through this portion of the graph that passes through it twice -$\displaystyle y = \frac{1}{2}$ would work. This half of the graph alone proves that $\displaystyle g^{-1}$ does not exist.

From Statement 1, the difference of two consecutive terms $\displaystyle a_{8}$ and $\displaystyle a_{9}$ is $\displaystyle a_{9} - a_{8 } = 72 - 65 = 7$; since, in an arithmetic sequence,

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

we can substitute $\displaystyle n = 9, d=7$ and find $\displaystyle a_{1}$.

$\displaystyle 72= a_{1} + (9-1) \cdot 7$

$\displaystyle 72= a_{1} +8 \cdot 7$

$\displaystyle 72= a_{1} + 56$

$\displaystyle a_{1} = 72 - 56 = 16$

However, Statement 2 alone gives insufficient information helpful in finding $\displaystyle a_{1}$; for example, the sequences 

$\displaystyle 1, 8, 15,...$

and 

$\displaystyle 2, 9, 16,...$

have the characteristic that $\displaystyle a_{2} +7 = a_{3}$ - but $\displaystyle a_{1}$ differs between them.

A function $\displaystyle f$ has an inverse if and only if, if $\displaystyle f(a) = f(b)$, then $\displaystyle a=b$, or, equivalently, if $\displaystyle a \ne b$, then $\displaystyle f(a) \ne f(b)$.

Assume Statement 2 alone. If $\displaystyle \left \{ 1, 2, 3, 4, 5 \right \}$ is the entire domain, then $\displaystyle f (a)$ cannot exist for any value of $\displaystyle a$ not in that set. Also, it can be seen that for each $\displaystyle a, b \in \left \{ 1, 2, 3, 4, 5 \right \}$ such that $\displaystyle a \ne b$, $\displaystyle f(a) \ne f(b)$. Therefore, $\displaystyle f$ has an inverse.

Assume Statement 1 alone. We show that the question of whether $\displaystyle f$ has an inverse cannot be answered by taking two cases.

Case 1: $\displaystyle \left \{ 1, 2, 3, 4, 5 \right \}$ is the entire domain. If this is true, then the range is $\displaystyle \left \{6, 7, 8, 9, 10 \right \}$, and the situation described in Statement 2 exists; consequently, $\displaystyle f$ has an inverse.

Case 2: $\displaystyle \left \{ 1, 2, 3, 4, 5 , 6 \right \}$ is the domain, and $\displaystyle f(6) = 6$. The range is still the set $\displaystyle \left \{6, 7, 8, 9, 10 \right \}$. However, $\displaystyle f(3) = f(6) = 6$, so there exists $\displaystyle a,b$ in the range such that $\displaystyle a \ne b$, but $\displaystyle f(a) = f(b)$. This means that $\displaystyle f$ does not have an inverse.

A function $\displaystyle f$ has an inverse if and only if, if $\displaystyle f(a) = f(b)$, then $\displaystyle a=b$, or, equivalently, if $\displaystyle a \ne b$, then $\displaystyle f(a) \ne f(b)$.

If Statement 1 alone is assumed, then this condition is known to not be true, since  $\displaystyle f(1)=f(6)= 7$. Therefore, $\displaystyle f$ does not have an inverse.

If Statement 2 alone is assumed, since no two values $\displaystyle a$ and $\displaystyle b$ are known such that $\displaystyle a \ne b$ and $\displaystyle f(a) = f(b)$, it is possible for $\displaystyle f$ to have an inverse. However, there may or may not be other values in the domain of $\displaystyle f$, any of which may be paired with range elements in the set $\displaystyle \left \{6, 7, 8, 9, 10 \right \}$. Therefore, Statement 2 does not resolve the issue of whether $\displaystyle f$ has an inverse.

Assume Statement 1 alone. Since $\displaystyle f$ is even, then by definition, for each $\displaystyle c$ in its domain, 

$\displaystyle f (c) = f (-c)$. 

Specifically, 

$\displaystyle f (4) = f (-4)$

and

$\displaystyle [f(4)+f(-4)] \cdot [g(4)+g(-4)] = [f(4)+f(4)] \cdot [g(4)+g(-4)] = 2 f(4) \cdot [g(4)+g(-4)]$.

Without further information, this expression cannot be evaluated.

Assume Statement 2 alone. Since $\displaystyle g$ is odd, then by definition, for each $\displaystyle c$ in its domain, 

$\displaystyle g (c) = - g (-c)$.

Specifically,

$\displaystyle g(4) = -g (-4)$, 

and

$\displaystyle [f(4)+f(-4)] \cdot [g(4)+g(-4)] = [f(4)+f(4)] \cdot [g(4)-g(-4)] = 2 f(4) \cdot 0 = 0$.

Assume both statements are true. A function $\displaystyle f$ is odd if, for all $\displaystyle c$ in its domain, $\displaystyle f(c) = -f(-c)$, and even if, for all $\displaystyle c$ in its domain, $\displaystyle f(c) = f(-c)$. We show that knowing that neither $\displaystyle f$ nor $\displaystyle g$ is odd or even is insufficient to answer the question of whether $\displaystyle g$ is odd, even, or neither by examining two scenarios.

Case 1: $\displaystyle f (x)= x+1$ and $\displaystyle g (x) = x-1$.

$\displaystyle f (1) = 1+1 = 2$

$\displaystyle f (-1) = -1+1 = 0$

Since there exists at least one value $\displaystyle c$ for which neither $\displaystyle f(c) = f(-c)$ nor $\displaystyle f(c) = -f(-c)$, $\displaystyle f$ is neither odd nor even.

By a similar argument, $\displaystyle g$ can be shown to be neither odd nor even.

However, 

$\displaystyle (fg) (x) = f(x)\cdot g(x) = (x+1)(x-1) = x^{2}-1$

and, for all $\displaystyle c$ in the domain,

$\displaystyle (fg) (-c) = (-c)^{2}-1 = c^{2} - 1 = (fg) (c)$,

making $\displaystyle fg$ even.

Case 2: $\displaystyle f (x)= x+1$ and $\displaystyle g (x) = x-2$.

Again, $\displaystyle f$ is neither even nor odd, and $\displaystyle g$ can be similarly demonstrated to be neither as well.

$\displaystyle (fg) (x) = f(x)\cdot g(x) = (x+1)(x-2)$

$\displaystyle (fg) (2) = (2+1)(2-2) = 3 \cdot 0 = 0$

$\displaystyle (fg) (-2) = (-2+1)(-2-2) = -1 \cdot (-4) = 4$

Since there is at least one value $\displaystyle c$ in the domain of $\displaystyle fg$ such that $\displaystyle fg(-c) \ne fg (c)$, $\displaystyle fg$ is neither odd nor even. 

$\displaystyle (f+g) (0) = f(0) + g(0)$, so we need to find the values of both $\displaystyle f(0)$ and $\displaystyle g(0)$ in order to answer this question.

Assume Statement 1 alone. By defintion of an odd function, from Statement 2, for every $\displaystyle a$ in the domain of $\displaystyle g$, $\displaystyle f(a) = -f(a)$. In specific, setting $\displaystyle a = 0$, 

$\displaystyle f(0) = -f(0)$.

The only number whose opposite is itself is 0, so 

$\displaystyle f(0) = 0$, 

and it follows that 

$\displaystyle (f+g) (0) = 0 + g(0) = g(0)$.

However, we have no way of knowing the value of $\displaystyle g(0)$, so the expression cannot be evaluated.

By a similar argument, if Statement 2 alone is assumed, $\displaystyle (f+g) (0) = f(0) + 0 = f(0)$, but, since $\displaystyle f(0)$ is unknown, the expression cannot be evaluated.

Now assume both statements. It follows that $\displaystyle f(0) = g(0) = 0$, and 

$\displaystyle (f+g) (0) = f(0) + g(0) = 0+0 = 0$.

$\displaystyle (f + g) (4) = f(4)+ g(4) = 9 + g(4)$, so to answer the question, it is necssary and sufficient to evaluate $\displaystyle g(4)$. Neither statement alone gives us this value. 

However, assume both statements to be true. By defintion of an odd function, from Statement 2, for every $\displaystyle a$ in the domain of $\displaystyle g$, $\displaystyle g(a) = -g(-a)$, so, in specific, $\displaystyle g(4) = -g(-4)$. From Statement 1, however, $\displaystyle g(4) = g(-4)$. This means that $\displaystyle g(4) = -g( 4)$, and $\displaystyle g(4)$ , being equal to its own opposite, must be equal to 0. Therefore,

$\displaystyle (f + g) (4) = 9 + g(4) = 9 + 0 = 9$.