✔ 最佳答案
1)
S3
= 3 x (a1 + a3) / 2
= 3 x [(2 x 1 + 1) + (2 x 3 + 1)] / 2
= 15
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2)
若「前 n 項」是「首 n 項和」之誤,則 S3
= 2 x 3 + 1
= 7
若「前 n 項」是「第 n 項」之誤,則與第 1 題相同。S3
= 15
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3)
a5 + a6 + ...... + a15
= S15 - S4
= 15 x (7 x 15 - 3) - 4 x (7 x 4 - 3)
= 1430
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4)
Sn = n[2a1 + (n - 1)d] / 2
S29
= 29 x [2a1 + (29 - 1)d] / 2
= 29 x [2a1 + 28)d] / 2
= 29 x [a1 + 14d]
= 29 x [a1 + (15 - 1)d]
= 29 x a15
= 29 x (-43)
= 1247
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5)
a6 :
a1 + (6 - 1)d = 13
2a1 + 10d = 26 ...... [1]
S10 :
10 x [2a1 + (10 - 1)d] / 2 = 120
5 x (2a1 + 9d) = 120
2a1 + 9d = 24 ...... [2]
[1] - [2] :
d = 2
把 d = 2 代入 [2] 中:
2a1 + 9 x 2 = 24
a1 = 3
a21
= 3 + (21 - 1) x 2
= 43