For an arithmetic sequence with first term 50 and common difference $- 4$, let $S _ { n }$ denote the sum of the first $n$ terms. What is the value of the natural number $m$ that maximizes $\sum _ { k = m } ^ { m + 4 } S _ { k }$? [4 points]
(1) 8
(2) 9
(3) 10
(4) 11
(5) 12

If the sum of the series $20 + 19 \frac { 3 } { 5 } + 19 \frac { 1 } { 5 } + 18 \frac { 4 } { 5 } + \ldots\ldots\ldots$ up to $n ^ { \text {th} }$ term is 488 and the $n ^ { \text {th} }$ term is negative, then :
(1) $n ^ { \text {th} }$ term is $- 4 \frac { 2 } { 5 }$
(2) $n = 41$
(3) $n ^ { \text {th} }$ term is - 4
(4) $n = 60$

Let $f : R \rightarrow R$ be such that for all $x \in R$, $\left( 2 ^ { 1 + x } + 2 ^ { 1 - x } \right)$, $f ( x )$ and $\left( 3 ^ { x } + 3 ^ { - x } \right)$ are in A.P., then the minimum value of $f ( x )$ is
(1) 2
(2) 3
(3) 0
(4) 4

Let $a_1, a_2, a_3, \ldots$ be an A.P. If $a_7 = 3$, the product $(a_1 a_4)$ is minimum and the sum of its first $n$ terms is zero then $n! - 4a_{n(n+2)}$ is equal to
(1) $\frac{381}{4}$
(2) 9
(3) $\frac{33}{4}$
(4) 24

In an A.P., the sixth term $\mathbf { a } _ { 6 } = 2$. If the $\mathbf { a } _ { 1 } \mathbf { a } _ { 4 } \mathbf { a } _ { 5 }$ is the greatest, then the common difference of the A.P., is equal to
(1) $\frac { 3 } { 2 }$
(2) $\frac { 8 } { 5 }$
(3) $\frac { 2 } { 3 }$
(4) $\frac { 5 } { 8 }$

Let $\{a_n\}$ be a sequence such that the sum $S_n$ of the terms from the first term to the $n$-th term is $$S_n = \frac{n^2 - 17n}{4},$$ and let $\{b_n\}$ be the sequence defined by $$b_n = a_n \cdot a_{n+5} \quad (n = 1, 2, 3, \cdots)$$
(1) For $\mathbf{A}$ $\sim$ $\mathbf{C}$ in the following sentences, choose the correct answer from among choices (0) $\sim$ (9) below.
Let us find the sum $T_n$ of the terms of sequence $\{b_n\}$ from the first term to the $n$-th term.
Since $a_n = \mathbf{A}$, we have $b_n = \mathbf{B}$. Hence we obtain $$T_n = \mathbf{C}.$$
(0) $\frac{n-7}{2}$
(1) $\frac{n-9}{2}$
(2) $\frac{n-11}{2}$
(3) $\frac{n^2 - 12n + 27}{4}$
(4) $\frac{n^2 - 13n + 36}{4}$
(5) $\frac{n^2 - 14n + 45}{4}$ (6) $\frac{n(n^2 - 17n + 83)}{12}$ (7) $\frac{n(n^2 - 17n + 89)}{12}$ (8) $\frac{n(n^2 - 18n + 83)}{12}$ (9) $\frac{n(n^2 - 18n + 89)}{12}$
(2) Next, let us find the minimum value of $T_n$.
When $n \leqq \mathbf{D}$ or $\mathbf{EF} \leqq n$, we see that $b_n > 0$. On the other hand, when $\mathbf{G} \leqq n \leqq \mathbf{H}$, we see that $b_n < 0$.
Hence $T_n$ is minimized at $n = \mathbf{I}$, $n = \mathbf{J}$ and $n = \mathbf{K}$, and its minimum value is $\mathbf{L}$. (Answer in the order such that $\mathbf{I} < \mathbf{J} < \mathbf{K}$.)