palmerconniba的部落格

1.
a2 = 52 = a1 + d -----(1)
a15 = 13 = a1 + 14d -----(2)
a20 = a1 + 19d
(2) - (1) = 13 - 52 = -39 = 13d
d = -3
a1 = 55
a20 = 55 + (-3) * 19 = -2

2.
a3 = 9 = a1 + 2d -----(1)
a5 = 25 = a1 + 4d -----(2)
a12 = a1 + 11d
(2) - (1) = 2d = 16
d = 8
a1 = -7
a12 = (-7) + 8*11 = 81

3.
(A)
a1,a2間公差顯然與a2,a3間公差不一樣
(B)
a1,a2間公差顯然與a2,a3間公差不一樣
(C)
a1,a2間公差顯然與a2,a3間公差不一樣
(D)
公差為10

4.
a5 - a4 = (a1 + 4d) - (a1 + 3d) = d = 6
a44 - a55 = (a1 + 43d) - (a1 + 54d) = -11d
= -66

5.
首項為20，末項為90，公差為7
20 + 7x = 90 (a1 + 7*x = a末)
x = 10 (代表需加10次7才可成為90)
等差數列中，(舉例) a1 到 a3，需要加2次公差才可以(a1到a2一次，a2到a3一次)
故此題答案為 x - 1 = 9

6.
a1 = 5/8 ; a末 = 12 + 5/8
(中間插入15個數>>代表 a1 到 a末 需加16次公差)
a1 + 16d = 5/8 + 16d = 12 + 5/8
d = 3/4
a9(插入的第八個數) = a1 + 8d = 5/8 + (3/4)*8 = 5/8 + 6
= 6又5/8

7.
a1 = -指揮棒高雄鳳山台北松山新莊泰山阿囉哈led總匯 ; a3 = -97
d = 2
an = a1 + 2 * n > 0
(-指揮棒高雄鳳山台北松山新莊泰山阿囉哈led總匯) + 2 * n > 0
2 * n > 指揮棒高雄鳳山台北松山新莊泰山阿囉哈led總匯
n > 50.5
n = 50

8.
已知 a1,a2,a3 三數成等差數列
則 a1 + a3 = a*a2
(2a-7) + (6a-21) = 2*中項
8a - 28 = 2*中項
中項 = 4a - 14