急!F6 Arithmetic Sequences 1q13

22.
(a)
The first term, a = 1
The common difference, d = -1 - 1 = -3 - (-2) = -2

The sum of the first 20 terms, S20
= 20 x (2a + 19d) / 2
= 20 x [2 x 1 + 19 x (-2)] / 2
= -360

(b)
The sum of the 21st term to the 40th term
= S40 - S21
= [40 x (2a + 39d) / 2] - (-360)
= [40 x (2 x 1 + 39 x (-2)) / 2] + 360
= -1160


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25.
(a)
The required sum
= (53 + 63 + 73 + ...... + 393)

It is sum of an arithmetic sequence with
first term = 53
common difference = 10
number of terms = [(393 - 53) / 2] + 1 = 35
last term = 393

The required sum
= 35 x (53 + 393) / 2
= 7805

(b)
52 + 62 + 72 + ...... + 392 is the sum of an arithmetic sequence with
first term = 52
common difference = 10
number of terms = [(392 - 52) / 2] + 1 = 35
last term = 392

52 + 62 + 72 + ...... + 392
= 35 x (52 + 392) / 2
= 7770

57 + 67 + 77 + ...... + 397 is the sum of an arithmetic sequence with
first term = 57
common difference = 10
number of terms = [(397 - 57) / 2] + 1 = 35
last term = 397

57 + 67 + 77 + ...... + 397
= 35 x (57 + 397) / 2
= 7945

The required sum
= 7770 + 7945
= 15715

2013-03-31 20:10:00 補充：
25. (b) Alternative method

= (52 + 62 + ... + 392) + (57 + 67 + ... + 397)
= [(53 - 1) + (63 - 1) + ... + (393 - 1)] + [(53 + 4) + (63 + 4) + ... + (393 + 4)]
= [(53 + 63 + ... + 393) - 1x35] + [(53 + 63 + ... + 393) + 4 x 35]
= 2 x (53 + 63 + ... + 393) + 3 x 35
= 2 x 7805 + 105
= 25715

2013-03-31 20:11:12 補充：
A typo :
The answer of the alternative method should be 15715

2013-03-31 20:11:37 補充：
25. (b) Alternative method

= (52 + 62 + ... + 392) + (57 + 67 + ... + 397)
= [(53 - 1) + (63 - 1) + ... + (393 - 1)] + [(53 + 4) + (63 + 4) + ... + (393 + 4)]
= [(53 + 63 + ... + 393) - 1x35] + [(53 + 63 + ... + 393) + 4 x 35]
= 2 x (53 + 63 + ... + 393) + 3 x 35
= 2 x 7805 + 105
= 15715