arithmetic progression/sequence question help?

[匿名]

2011-07-25 7:15 am

1) The nth term of an arithmetic progression is 32 and its first term is 2. If the sum of the first n term is 357, find the value of n. If the smallest term in the series which exceeds 100 is the kth term, find the value of k.

2) The sum of the first n terms of an arithmetic progression is given by Sn = pn + qn2 . Given that S3 = 6 and S5 = 11,
a) find the values of p and q,
b) deduce, or otherwise, an expression for the nth term and the term and the value of its common difference.

Please explain to me step by step
Thank you very much

回答 (1)

用戶57799

2011-07-25 7:35 am

✔ 最佳答案

1) 32 = 2+(n-1)*d
(n-1)*d = 30
357 = (4+(n-1)*d)*n/2
714 = (4+30)n =34n
a)n = 21
b)32 = 2+20d
d = 3/2
nth term = 2+(n-1)*3/2
100 = 2+(k-1)*3/2
200 = 4 +3k-3
k = 66.33
so k = 67............Answer
and the 67th term is 101
2) a)6 = 3p + 9q times 5 makes 30=15p+45q
11=5p+25q times 3 makes.... 33=15p+75q now subtract equations
3 = 30q
q = 1/10
p = 17/10
b) a = first term = 17/10 + 1/10 = 18/10
sum to n terms = (2a+(n-1)d)n/2
6 = (36/10+2d)*3/2
40=36+20d
Common difference d = 1/5 ........Answer
nth term = 18/10+(n-1)/5 = (18+2n-2) /10 = (n+8)/5.........Answer
The series is 9/5,10/5,11/5,12/5.13/5.......(n+8)/5

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