절댓값 합 조건 등차수열

[출제의도] 등차수열의 성질을 활용하여 문제 해결하기 등차수열 $\left\{ a _{n} \right\}$의 첫째항을 $a$, 공차를 $d$라 하자. 조건 (가)에 의하여 $\displaystyle \sum\limits _{n = 1} ^{2 m - 1} a _{n} = \frac{\left( 2 m - 1 \right) \left\{ 2 a + \left( 2 m - 2 \right) d \right\}}{2}$ $= \left( 2 m - 1 \right) \left\{ a + \left( m - 1 \right) d \right\} = 0$ $2 m - 1 > 0$이므로 $a + \left( m - 1 \right) d = 0$ 즉, $a _{m} = 0$ 조건 (나)에 의하여 $\displaystyle \sum\limits _{n = 1} ^{15} a _{n} \ne \sum\limits _{n = 1} ^{15} \left| a _{n} \right|$이고 $\displaystyle \sum\limits _{n = 1} ^{15} a _{n} > 0$이므로 $a _{m} = 0$을 만족시키는 $m$의 범위는 ${m} \leq 7$

$\displaystyle \sum\limits _{n = 1} ^{2 m - 1} a _{n} = 0$이므로 $\displaystyle \sum\limits _{n = 1} ^{15} a _{n} = \frac{\left( 16 - 2 m \right) \left\{ 2 md + \left( 15 - 2 m \right) d \right\}}{2}$ $= 15 \left( 8 - m \right) d = 45$ 즉, $\left( 8 - m \right) d = 3$ $\displaystyle \sum\limits _{n = 1} ^{15} \left| a _{n} \right|$$\displaystyle = 2 \times \frac{\left( m - 1 \right) \left\{ d + \left( m - 1 \right) d \right\}}{2} + 15 \left( 8 - m \right) d$ $= \left( m - 1 \right) md + 45 = 90$ $\displaystyle 2 \sum\limits _{n = 1} ^{15} a _{n} = \sum\limits _{n = 1} ^{15} \left| a _{n} \right|$이므로 $15 \left( 8 - m \right) d = \left( m - 1 \right) md$ $m ^{2} + 14 m - 120 = 0$ $\left( m - 6 \right) \left( m + 20 \right) = 0$ $m$은 자연수이므로 $m = 6$ $\displaystyle \sum\limits _{n = 1} ^{15} a _{n} = 30 d = 45$ 즉, $\displaystyle d = \frac{3}{2}$ 따라서 $a _{14} = a _{6} + 8 d = 0 + 12 = 12$