First, multiple choice questions
1. If Sn is the sum of the first n items of the sequence {an}, it is ().
A. geometric progression, but not geometric progression B. geometric progression, but not geometric progression.
C. arithmetic progression, and geometric progression D. Neither geometric progression nor arithmetic progression.
2. In the process of culture, a bacterium divides every 20 minutes (one divides into two), and after 3 hours, the bacterium can reproduce from 1 to ().
a . 5 1 1 b . 5 12 c . 1023d . 1024。
3. In arithmetic progression {a n}, () is known.
A.48 B.49 C.50 D.5 1
4. It is known that {an} is a geometric series, and an > 0, a2a4+a3+a5+a4a6 = 25, then the value of a3+a5 is equal to ().
10 C. 15 D.20
5. The first term of geometric series {An} a 1 = 1, and the common ratio q≠ 1. If A 1, a2 and a3 are arithmetic progression's 1, 2 and 5 in turn, then Q is equal to ().
A. 2b. 3c-3d. 3 or -3
6. The sum of the first three terms of a geometric series {{an}} is equal to 3 times of the first term, so the common ratio of this geometric series is ().
A.-2b. 1 c.-2 or 1 D.2 or-1
7. Given a arithmetic progression whose first term consists of four roots of the equation, then ().
65438 AD+0 BC
8. In the sequence {an}, it is known that S 1 = 1, S2=2, Sn+1-3sn+2sn-1= 0 (n ∈ n *), so this sequence is ().
A. arithmetic progression B. Geometric sequence
C. From the second term, arithmetic progression D. From the second term, geometric series
9. If the sum of the first few terms in the geometric series is 54 and the sum of the first few terms is 60, then the sum of the first few terms is ().
AD 66-64
10. Let the tolerance of arithmetic progression {an} be d, and if the first n terms are sn =-N2, then ().
A. an =2n- 1, d=-2.
C.an=-2n+ 1，d=-2 D.an=-2n+ 1，d=2
The general formula of 1 1. series {an} is a n = (n∈N*). If the sum of the first n items is 10, the number of items is ().
a . 1 1 b . 99 c . 120d . 12 1
1 2. On July1day, 2000, someone went to the bank to deposit one yuan, which was fixed for one year. It is planned to withdraw the principal and interest of the due deposit in July 1 and 200 1 and deposit it for another one-year term. After that, he will withdraw money from the bank in the same way every July 1 day. Then when he withdrew all the deposits and principal and interest on July 1 2005, the money he withdrew was ().
A. A (1+R) 4 yuan B (1+R) 5 yuan
C.A. (1+R) 6 yuan D. [( 1+R) 6-( 1+R)] Yuan.
Second, fill in the blanks:
13. Let {an} be a geometric series whose common ratio is q, and Sn is the sum of its first n terms. If {sn} is arithmetic progression,
Then q =
14. Let the sequence satisfy when.
15. The sum of the first n terms of the sequence Sn = 3N2+N+ 1, then the general term formula A N = _ _.
16. In arithmetic progression, when must be a constant sequence. But in geometric series, for some positive integers, an example of non-constant series is _ _ _ _ _.
Third, answer questions:
17. Known: arithmetic progression {}, = 14, the sum of the former 10.
( 1);
(2) Arrange the second item, the fourth item, …, and the first item in {} into a new series in the original order, and find the sum of the previous items in this series.
18. Find the sum of the following series:
( 1) ;
(2)
19. The sequence {an} satisfies a 1= 1, and an = An- 1+ 1 (n ≥ 2).
(1) If BN = an-2, verify that {BN} is a geometric series;
(2) Find the general formula of {{an}}.
20. A fishing company bought a fishing boat for 980,000 yuan at the beginning of the year. In the first year, the expenses were 6.5438+0.2 million yuan, and then increased by 40 thousand yuan every year, and the annual fishing income was 500 thousand yuan.
(1) When did you start making profits?
(2) After several years, there are two treatment schemes:
(3) When the annual average profit is the largest, sell the fishing boat for 260,000 yuan;
(4) When the total net profit is maximum, the fishing boat sells 80,000 yuan.
Ask which scheme is cost-effective.
2 1. The known sequence is arithmetic progression.
(1) Find the general term formula of the sequence;
(2) The formula for finding the sum of the first n items in the series.
22. There are two plans for selling houses launched by a real estate company: one is the installment plan, which requires buyers to pay 30,000 yuan in the first year, and then pay 8,000 yuan every year for ten consecutive years from the second year; Another option is a one-time payment with a preferential price of 90,000 yuan. If a homebuyer has 90,000 yuan in cash to buy a house, which option should be used to buy a house, considering that the annual return rate of another investment is 5%? Please explain the reason. (Reference data1.059 ≈1.551.051.628)
Reference answer
First, multiple-choice questions: BBCAB CCDDC CD
Fill in the blanks: 13. 1. 14.
15 ... 16, and are both odd or even numbers.
Third, answer questions:
17. Analysis: (1) by VIII
pass by
(1) Set the new series as {},
18. Analysis: (1)
(The method used in this question is called "split term method", and the general term is expressed in the form of an = f (n+ 1)-f (n))
(2) The general term is "equal difference × equal ratio",
19. Analysis: (1) An-2 = (an- 1+0)
That is, (n≥2)
∴ {BN} is a geometric series with-1 as the first term, the common ratio.
(2) bn = (- 1) () n- 1, that is, an-2 =-() n- 1.
∴an=2-( )n- 1
20. Analysis: (1) According to the title, the annual cost is arithmetic progression with 12 as the first term and 4 as the tolerance. Suppose the relationship between net income and years is,
∴ ,
Profit > 0, ∴,
Solution:
And n∈N, ∴ n=3,4, …, 17, ∴ when n = 3, it starts to make profits in the third year;
(1) (i) Average annual income =
∫≥, take "=" if and only if n=7,
∴≤ 40-2× 14 = 12 (ten thousand yuan), that is, the average annual income, and the total income is 12×7+26 = 165438+ ten thousand yuan, at this time n = 7.
(ii) When
The total income is 102+8 = 165438+ ten thousand yuan, and n= 10. Comparing the two schemes, the total income is 165438+ 10,000 yuan, but the first scheme needs 7 years and the second scheme needs 10 years.
2 1. Analysis: If the tolerance of the series is 0, then again
So (Ⅱ) solution: get the order.
① ②
When ① subtracts ②, you get
therefore
At the right time, at the right time,
When,
22. Analysis: If the payment is made by installments, it depends on whether there is any balance after the payment in the first year 1 1, and let the balance in the nth year after the first payment be an.
∫a 1 =(9-3)×( 1+0.5%)-0.8 = 6× 1.05-0.8
a2 =(6× 1.05-0.8)× 1.05-0.8 = 6× 1.052-0.8×( 1+ 1.05)
……
a 10 = 6× 1.05 10-0.8( 1+ 1.05+…+ 1.059)
=6× 1.05 10-0.8×
=6× 1.05 10- 16×( 1.05 10- 1)
= 16- 10× 1.05 10
≈ 16- 16.28 =-0.28 (ten thousand yuan)
So a one-time payment is cost-effective.