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Sequence - Practice Questions (38)

Question 1: 1. The equivariant middle term of $2 + \sqrt { 3 }$ and $2 - \sqrt { 3 }$ is

1. The equivariant middle term of $2 + \sqrt { 3 }$ and $2 - \sqrt { 3 }$ is

  • A. A. 1
  • B. B. - 1
  • C. C. $\pm 1$
  • D. D. 2

Answer: C

Solution: Let the middle term of the equality ratio be $a$, then $a ^ { 2 } = ( 2 + \sqrt { 3 } ) ( 2 - \sqrt { 3 } ) = 1$, then $a = \pm 1$.

Question 2: 2. If $\left\{ a _ { n } \right\}$ is known to be an equivariant series, and if $a _ { 4 } = 15$ , t...

2. If $\left\{ a _ { n } \right\}$ is known to be an equivariant series, and if $a _ { 4 } = 15$ , then the sum of the first 7 terms of it is

  • A. A. 120
  • B. B. 115
  • C. C. 110
  • D. D. 105

Answer: D

Solution: Since the isomorphic series $\left\{ a _ { n } \right\}$ in $a _ { 4 } = 15$, $a _ { 4 } = 15$ so $S _ { 7 } = \frac { 7 } { 2 } \left( a _ { 1 } + a _ { 7 } \right) = \frac { 7 } { 2 } \cdot 2 a _ { 4 } = 7 a _ { 4 } = 7 \times 15 = 105$.

Question 3: 3. The real numbers $m , 3,2$ are known to form an equidistant series, then the centroid of the coni...

3. The real numbers $m , 3,2$ are known to form an equidistant series, then the centroid of the conic curve $\frac { x ^ { 2 } } { m } + y ^ { 2 } = 1$ is

  • A. A. $\frac { \sqrt { 5 } } { 2 }$
  • B. B. $\sqrt { 5 }$
  • C. C. $\frac { \sqrt { 3 } } { 2 }$
  • D. D. $\sqrt { 3 }$

Answer: C

Solution: SOLUTION: Since $m , 3,2$ forms an equidistant series, $\therefore 2 + m = 6 , \therefore m = 4$, the equation of the ellipse is $\frac { x ^ { 2 } } { 4 } + y ^ { 2 } = 1$, $\frac { x ^ { 2 } } { 4 } + y ^ { 2 } = 1$ $\therefore a ^ { 2 } = 4 , b ^ { 2 } = 1 , \therefore a = 2 , c = \sqrt { 3 } , \therefore e = \frac { \sqrt { 3 } } { 2 }$.

Question 4: 4. It is known that the sum of the first $n$ terms of the isomorphic series $\left\{ a _ { n } \righ...

4. It is known that the sum of the first $n$ terms of the isomorphic series $\left\{ a _ { n } \right\}$ is $S _ { n } , ~ a _ { 1 } \neq 0 , ~ S _ { 8 } = 0$, then

  • A. A. $\left\{ a _ { n } \right\}$ must be an increasing number series
  • B. B. $\left\{ a _ { n } \right\}$ must be a decreasing number series
  • C. C. $a _ { 4 } + a _ { 5 } = 0$
  • D. D. $a _ { 3 } + a _ { 5 } = 0$

Answer: C

Solution: $S _ { 8 } = \frac { \left( a _ { 1 } + a _ { 8 } \right) \times 8 } { 2 } = 4 \left( a _ { 1 } + a _ { 8 } \right) = 0$ $\therefore ^ { a _ { 1 } + a _ { 8 } = 0 }$. $a _ { 4 } + a _ { 5 } = 0$ according to the properties of an equivariant series.

Question 5: 5. In the isoperimetric series $\left\{ a _ { n } \right\}$, $a _ { 1 } = \frac { 9 } { 8 } , q = \f...

5. In the isoperimetric series $\left\{ a _ { n } \right\}$, $a _ { 1 } = \frac { 9 } { 8 } , q = \frac { 2 } { 3 } , S _ { n } = \frac { 19 } { 8 }$, then $n =$

  • A. A. 3
  • B. B. 4
  • C. C. 5
  • D. D. 6

Answer: A

Solution: From the question, $S _ { n } = \frac { a _ { 1 } \left( 1 - q ^ { n } \right) } { 1 - q } = \frac { \frac { 9 } { 8 } \left( 1 - \left( \frac { 2 } { 3 } \right) ^ { n } \right) } { 1 - \left( \frac { 2 } { 3 } \right) } = \frac { 19 } { 8 }$, solve $\quad$.

Question 6: 6. It is known that the common ratio of the isometric series $\left\{ a _ { n } \right\}$ in $a _ { ...

6. It is known that the common ratio of the isometric series $\left\{ a _ { n } \right\}$ in $a _ { 2 } = - 2 , a _ { 5 } = 16$ is

  • A. A. $\pm 2$
  • B. B. 2
  • C. C. - 2
  • D. D. 3

Answer: C

Solution: Let the common ratio be $q$ and $q ^ { 3 } = \frac { a _ { 5 } } { a _ { 2 } } = \frac { 16 } { - 2 } = - 8 , q = - 2$.

Question 7: 7. In the isomorphic series $\left\{ a _ { n } \right\}$, it is known that $a _ { 2 } + a _ { 6 } = ...

7. In the isomorphic series $\left\{ a _ { n } \right\}$, it is known that $a _ { 2 } + a _ { 6 } = 18$, then $a _ { 4 } = ($

  • A. A. 9
  • B. B. 8
  • C. C. 81
  • D. D. 63

Answer: A

Solution: From the properties of the isomorphic series $a _ { 2 } + a _ { 6 } = 2 a _ { 4 }$ , $a _ { 2 } + a _ { 6 } = 2 a _ { 4 }$ $\because a _ { 2 } + a _ { 6 } = 18$ $\therefore 2 a _ { 4 } = 18$ INLINE_FORMULA_1]] $\therefore 2 a _ { 4 } = 18$ and $a _ { 4 } = 9$, so choose A . This question mainly examines the application of the properties of the isomorphic series, which is a simple question. If $m + m = q + p _ { \text {则 } } a _ { m } + a _ { n } = a _ { p } + a _ { q }$ in the isomorphic series $\left\{ a _ { n } \right\}$.

Question 8: 8. The sum of the first $\left\{ a _ { n } \right\}$ terms of the isomorphic series $n$ is $S _ { n ...

8. The sum of the first $\left\{ a _ { n } \right\}$ terms of the isomorphic series $n$ is $S _ { n }$, and if $a _ { 4 } + a _ { 6 } = 12$, then the value of $S _ { 9 }$ is

  • A. A. 36
  • B. B. 48
  • C. C. 54
  • D. D. 64

Answer: C

Solution: From the properties of the isomorphic series $\left\{ \mathrm { a } _ { \mathrm { n } } \right\}$, we have: $\mathrm { a } _ { 4 } + \mathrm { a } _ { 6 } = 12 = \mathrm { a } _ { 1 } + \mathrm { a } _ { 9 }$ . Then $\mathrm { S } _ { 9 } = \frac { 9 \left( a _ { 1 } + a _ { 9 } \right) } { 2 } = 9 \times \frac { 12 } { 2 } = 54$ .

Question 9: 9. $S n$ is known to be the sum of the first $n$ terms of the isomorphic series $\{ a n \}$ whose to...

9. $S n$ is known to be the sum of the first $n$ terms of the isomorphic series $\{ a n \}$ whose tolerance is not 0, $S _ { 9 } = 18$, $a m = 2$, and $m =$ ( ). INLINE_FORMULA_5]] ( )

  • A. A. 4
  • B. B. 5
  • C. C. 6
  • D. D. 7

Answer: B

Solution: SOLUTION: $S _ { 9 } = \frac { 9 \left( a _ { 1 } + a _ { 9 } \right) } { 2 } = 9 a _ { 5 } = 18 , \therefore a _ { 5 } = 2$ $\because a m = 2 , \therefore m = 5$ .

Question 10: 10. The middle term of the equivalence of $\sqrt { 3 } - 1$ and $\sqrt { 3 } + 1$ is ( )

10. The middle term of the equivalence of $\sqrt { 3 } - 1$ and $\sqrt { 3 } + 1$ is ( )

  • A. A. $\sqrt { 2 }$
  • B. B. $- \sqrt { 2 }$
  • C. C. $\pm \sqrt { 2 }$
  • D. D. $\pm \frac { \sqrt { 2 } } { 2 }$

Answer: C

Solution: The equivariant middle term of $\sqrt { 3 } - 1$ and $\sqrt { 3 } + 1$ is $\pm \sqrt { ( \sqrt { 3 } - 1 ) ( \sqrt { 3 } + 1 ) } = \pm \sqrt { 2 }$.

Question 11: 11. $\left\{ a _ { n } \right\}$ is an isometric series and $a _ { 7 } - 2 a _ { 4 } = - 1 , a _ { 3...

11. $\left\{ a _ { n } \right\}$ is an isometric series and $a _ { 7 } - 2 a _ { 4 } = - 1 , a _ { 3 } = 0$ , then the tolerance $d =$

  • A. A. - 2
  • B. B. $- \frac { 1 } { 2 }$
  • C. C. $\frac { 1 } { 2 }$
  • D. D. 2

Answer: B

Solution: Test analysis : $\because a _ { 7 } - 2 a _ { 4 } = - 1 \therefore a _ { 3 } + 4 d - 2 \left( a _ { 3 } + d \right) = - 1 \therefore 4 d - 2 d = - 1 \therefore d = - \frac { 1 } { 2 }$ Points : Equivariant series general formula

Question 12: 12. In the isomorphic series $\{ a n \}$, $a _ { 3 } + a _ { 5 } = 10$, then $a _ { 1 } + a _ { 7 }$...

12. In the isomorphic series $\{ a n \}$, $a _ { 3 } + a _ { 5 } = 10$, then $a _ { 1 } + a _ { 7 }$ is equal to ( )

  • A. A. 5
  • B. B. 8
  • C. C. 10
  • D. D. 14

Answer: C

Solution: $a _ { 1 } + a _ { 7 } = a _ { 3 } + a _ { 5 } = 10$.

Question 13: 13. It is known that the series $\left\{ a _ { n } \right\}$ in $a _ { 3 } = 2 , a _ { 7 } = 1$, and...

13. It is known that the series $\left\{ a _ { n } \right\}$ in $a _ { 3 } = 2 , a _ { 7 } = 1$, and the series $\left\{ \frac { 1 } { a _ { n } + 1 } \right\}$ is an isometric series, then $a _ { 11 } =$

  • A. A. $- \frac { 2 } { 5 }$
  • B. B. $\frac { 1 } { 2 }$
  • C. C. 5
  • D. D. $\frac { 2 } { 3 }$

Answer: B

Solution: 试题分析:数列 $\left\{ \frac { 1 } { a _ { n } + 1 } \right\}$ has the third term $\frac { 1 } { a _ { 3 } + 1 } = \frac { 1 } { 3 }$ and the seventh term $\frac { 1 } { a _ { 7 } + 1 } = \frac { 1 } { 2 }$, so the eleventh term is $\frac { 1 } { a _ { 7 } + 1 } = \frac { 1 } { 2 }$. item is $\frac { 1 } { a _ { 11 } + 1 } = \frac { 2 } { 3 } \therefore a _ { 11 } = \frac { 1 } { 2 }$, so the eleventh item is $\frac { 1 } { a _ { 11 } + 1 } = \frac { 2 } { 3 } \therefore a _ { 11 } = \frac { 1 } { 2 }$. Points: Equal variance series

Question 14: It is known that the first term $a _ { 1 } = 1$ of the positive isometric series $\left\{ a _ { n } ...

It is known that the first term $a _ { 1 } = 1$ of the positive isometric series $\left\{ a _ { n } \right\}$ and the sum of the first $n$ terms is $S _ { n }$, and $\mathrm { S } _ { 1 } , \mathrm {~S} _ { 2 } , S _ { 3 } - 2$ is an isotropic series, then $a _ { 4 } =$ is $a _ { 4 } =$. And $\mathrm { S } _ { 1 } , \mathrm {~S} _ { 2 } , S _ { 3 } - 2$ is an equal difference series, then $a _ { 4 } =$ ( ).

  • A. A. 8
  • B. B. $\frac { 1 } { 8 }$
  • C. C. 16
  • D. D. $\frac { 1 } { 16 }$

Answer: A

Solution: Let the common ratio of the isometric series $\left\{ a _ { n } \right\}$ be $q$ because $\mathrm { S } _ { 1 } , \mathrm {~S} _ { 2 } , \mathrm {~S} _ { 3 } - 2$ forms an isometric series. So $2 S _ { 2 } = S _ { 1 } + S _ { 3 } - 2$, so $2 \left( a _ { 1 } + a _ { 2 } \right) = a _ { 1 } + a _ { 1 } + a _ { 2 } + a _ { 3 } - 2$ So $a _ { 2 } = a _ { 3 } - 2$, which is $q = q ^ { 2 } - 2$, solves $q = 2$ or $q = - 1$. Since $a _ { n } > 0$, it is $q = 2$ and so $a _ { 4 } = a _ { 1 } q ^ { 3 } = 8$.

Question 15: 15.According to global skyscraper statistics, by 2019, Hefei City, Anhui Province, has 95 skyscraper...

15.According to global skyscraper statistics, by 2019, Hefei City, Anhui Province, has 95 skyscrapers in Chinese cities ranked 10th, ranked 15th globally, and the tallest building under construction in the Evergrande Center in Hefei is designed in the shape of "bamboo joints", which not only embodies the power of extraordinary, but also symbolizes the strong will to grow upward, but also signals future prosperity and prosperity. The shape of the tallest building under construction of Evergrande Center in Hefei is designed in the form of "bamboo joints", which not only reflects the extraordinary strength, but also symbolizes the strong will of upward growth, and moreover, it foretells the future prosperity and prosperity. It and the inheritance of thousands of years of "micro-culture" complement each other, after the completion of the world's top ten skyscrapers, if the building consists of nine sections of "bamboo", the top part of the four sections of 228 meters high, the bottom part of the three sections of 204 meters high, and each section of the height of the changes in height evenly (i.e., the height of each section from top to bottom into an equal series), then the total height of the skyscraper is ( )

  • A. A. 518 meters.
  • B. B. 558 meters
  • C. C. 588 meters
  • D. D. 668 meters

Answer: B

Solution: Let each section of the building from the top down in height form an isotropic series $\left\{ a _ { n } \right\}$ . Let the first term of the series be ${ } ^ { a }$ and the tolerance ${ } ^ { d }$ . From the question $S _ { 4 } = 228 , S _ { 9 } - S _ { 6 } = 204$ , $S _ { 4 } = 228 , S _ { 9 } - S _ { 6 } = 204$ $4 a _ { 1 } + 6 d = 228 \quad , 3 a _ { 1 } + 21 d = 204 ;$ The system of joint equations is solved by $a _ { 1 } = 54 , d = 2$. Therefore, $S _ { 9 } = 9 a _ { 1 } + 36 d = 54 \times 9 + 36 \times 2 = 558$ is obtained.

Question 16: 17. In $V A B C$, $a , ~ b , ~ c$ is the opposite side of angle $A , ~ B , ~ C$. If $a , ~ b , ~ c$ ...

17. In $V A B C$, $a , ~ b , ~ c$ is the opposite side of angle $A , ~ B , ~ C$. If $a , ~ b , ~ c$ forms an equipartite series and $a ^ { 2 } - c ^ { 2 } = ( a - b ) c$, then the size of $A$ is

  • A. A. $\frac { \pi } { 6 }$
  • B. B. $\frac { \pi } { 3 }$
  • C. C. $\frac { 2 \pi } { 3 }$
  • D. D. $\frac { 5 \pi } { 6 }$

Answer: B

Solution: From known, we have $b ^ { 2 } = a c$, from $a ^ { 2 } - c ^ { 2 } = ( a - b ) c$, we have $a ^ { 2 } - c ^ { 2 } = a c - b c$, so $a ^ { 2 } - c ^ { 2 } = b ^ { 2 } - b c$, we have $b c = b ^ { 2 } + c ^ { 2 } - a ^ { 2 }$. By the cosine theorem, we have $\cos A = \frac { b ^ { 2 } + c ^ { 2 } - a ^ { 2 } } { 2 b c } = \frac { b c } { 2 b c } = \frac { 1 } { 2 }$, and $A \in ( 0 , \pi )$, so $A = \frac { \pi } { 3 }$.

Question 17: 18. In the series $\left\{ a _ { n } \right\}$, $a _ { 1 } = 1 , a _ { 2 } = 2$, for $\forall n \in ...

18. In the series $\left\{ a _ { n } \right\}$, $a _ { 1 } = 1 , a _ { 2 } = 2$, for $\forall n \in \mathbf { N } ^ { * } , a _ { n + 2 } = \frac { 5 } { 2 } a _ { n + 1 } - \frac { 3 } { 2 } a _ { n }$, then $a _ { 2021 } =$

  • A. A. $2 \left( \frac { 3 } { 2 } \right) ^ { 2018 } - 1$
  • B. B. $2 \left( \frac { 3 } { 2 } \right) ^ { 2019 } - 1$
  • C. C. $2 \left( \frac { 3 } { 2 } \right) ^ { 2020 } - 1$
  • D. D. $2 \left( \frac { 3 } { 2 } \right) ^ { 2021 } - 1$

Answer: C

Solution: Solution: From $a _ { n + 2 } = \frac { 5 } { 2 } a _ { n + 1 } - \frac { 3 } { 2 } a _ { n }$ we get $a _ { n + 2 } - a _ { n + 1 } = \frac { 3 } { 2 } \left( a _ { n + 1 } - a _ { n } \right)$ $\therefore$ series $\left\{ a _ { n + 1 } - a _ { n } \right\}$ is an isometric series with $a _ { 2 } - a _ { 1 } = 1$ as the first term and $\frac { 3 } { 2 }$ as the common ratio. $\therefore a _ { n + 1 } - a _ { n } = \left( \frac { 3 } { 2 } \right) ^ { n - 1 } \left( n \in N ^ { * } \right)$ $\therefore$ when $n \geq 2$, $a _ { n } = \left( a _ { n } - a _ { n - 1 } \right) + \left( a _ { n - 1 } - a _ { n - 2 } \right) + \cdots + \left( a _ { 3 } - a _ { 2 } \right) + \left( a _ { 2 } - a _ { 1 } \right) + a _ { 1 }$ $= \left( \frac { 3 } { 2 } \right) ^ { n - 2 } + \left( \frac { 3 } { 2 } \right) ^ { n - 3 } + \cdots + \left( \frac { 3 } { 2 } \right) ^ { 1 } + \left( \frac { 3 } { 2 } \right) ^ { 0 } + 1$ $= \frac { 1 - \left( \frac { 3 } { 2 } \right) ^ { n - 1 } } { 1 - \frac { 3 } { 2 } } + 1$ $= 2 \left( \frac { 3 } { 2 } \right) ^ { n - 1 } - 1$ The $n = 1$ holds when $n = 1$ is examined. $\therefore a _ { n } = 2 \left( \frac { 3 } { 2 } \right) ^ { n - 1 } - 1$. $\therefore a _ { 2021 } = 2 \left( \frac { 3 } { 2 } \right) ^ { 2020 } - 1$.

Question 18: 19. The Chinese Remainder Theorem, also known as Sun Tzu's Theorem, is about division. The 2024 numb...

19. The Chinese Remainder Theorem, also known as Sun Tzu's Theorem, is about division. The 2024 numbers from 1 to 2024 that are divisible by 3 and 1 and divisible by 5 are listed in descending order as $\left\{ a _ { n } \right\}$, preceded by $n _ { \text {项和为 } } S _ { n }$ and $S _ { 20 } - a _ { 10 } =$, and $S _ { 20 } - a _ { 10 } =$.

  • A. A. 2130
  • B. B. 2734
  • C. C. 2820
  • D. D. 3019

Answer: B

Solution: The number divisible by 3 and 1 and the number divisible by 5 is the number divisible by 15 and 1. Arranged in order from smallest to largest in an isomorphic series with 1 as the first term and a tolerance of 15, then $a _ { n } = 1 + 15 \times ( n - 1 ) = 15 n - 14$, $a _ { n } = 1 + 15 \times ( n - 1 ) = 15 n - 14$ so $a _ { 10 } = 15 \times 10 - 14 = 136 , S _ { 20 } = 1 \times 20 + \frac { 20 \times 19 } { 2 } \times 15 = 2870$ . so $S _ { 20 } - a _ { 10 } = 2870 - 136 = 2734$.

Question 19: 20. The standard logarithmic visual acuity chart (pictured) uses the "five-point recording method", ...

20. The standard logarithmic visual acuity chart (pictured) uses the "five-point recording method", which is a unique way of recording visual acuity in China. Standard logarithmic visual acuity table rows of square "$E$" word vision mark, and from the visual acuity of 5.1 vision mark in the rows upward, each line "$E$ is $\sqrt [ 10 ] { 10 }$ times the length of the side of "$E$" in the row below it, and if the length of the side of the reticle with visual acuity 4.0 is ${ } ^ { a }$, then the length of the side of the reticle with visual acuity 4.9 is | Standardized Log Distance Visual Acuity Chart | | | :--- | :--- | | ![](/images/questions/sequence/image-001.jpg) | | | [IMAGE_9]] | [IMAGE_9]] | [IMAGE_9]] | | | | | E ${ } ^ { 43 }$ | | | The Government of the United States of America has been working on a number of initiatives to improve the quality of education. | The following is a summary of the work done by the organization. | m Em $\boldsymbol { \Xi } 4.8$ | | | E m Yin $\omega$ E $\mathrm { m } \equiv$ 5.0 | | | | 5.0 | 5.1 | 5.0 | | 5.1 | | 5.2 | | |

  • A. A. $10 ^ { \frac { 4 } { 5 } } a$
  • B. B. $10 ^ { \frac { 9 } { 10 } } a$
  • C. C. $10 ^ { - \frac { 4 } { 5 } } a$
  • D. D. $10 ^ { - \frac { 9 } { 10 } } a$

Answer: D

Solution: Let row $n$ be $a _ { n }$ and row $n - 1$ be $a _ { n - 1 } ( n \geq 2 )$. From the question, $a _ { n - 1 } = \sqrt [ 10 ] { 10 } a _ { n } ( n \geq 2 )$ , then $\frac { a _ { n } } { a _ { n - 1 } } = 10 ^ { - \frac { 1 } { 10 } } ( n \geq 2 )$ , then the series $\left\{ a _ { n } \right\}$ is the first term of $a$ and the common ratio is $10 ^ { - \frac { 1 } { 10 } }$ and the common ratio is $10 ^ { - \frac { 1 } { 10 } }$. So $a _ { 10 } = a \left( 10 ^ { - \frac { 1 } { 10 } } \right) ^ { 10 - 1 } = 10 ^ { - \frac { 9 } { 10 } } a$, the length of the side of the reticle with visual acuity 4.9 is $10 ^ { - \frac { 9 } { 10 } } a$, and the length of the side of the reticle with visual acuity 4.9 is $10 ^ { - \frac { 9 } { 10 } } a$.

Question 20: 21. Let the left and right foci of hyperbola $\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } ...

21. Let the left and right foci of hyperbola $\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > 0 , b > 0 )$ be $F _ { 1 } , F _ { 2 }$, if there exists a point $P$ on the right branch of the hyperbola such that $\left| P F _ { 2 } \right| , \left| P F _ { 1 } \right| , \left| F _ { 1 } F _ { 2 } \right|$ is an equidistant series, then the range of values of the centrality of this hyperbola is ( )

  • A. A. $[ 3 , + \infty )$
  • B. B. $( 1,3 ]$
  • C. C. $( 3 , + \infty )$
  • D. D. $( 1,3 )$

Answer: A

Solution: Let $\left| P F _ { 2 } \right| = m$ , then $\left| P F _ { 1 } \right| = m + 2 a , \left| F _ { 1 } F _ { 2 } \right| = m + 4 a = 2 c$ , so $m = 2 c - 4 a$ , the and $P$ is on the right branch, so $\left| P F _ { 2 } \right| \geq c - a$, which is $2 c - 4 a \geq c - a$, and so the eccentricity $e = \frac { c } { a } \geq 3$.

Question 21: The series $\left\{ a _ { n } \right\}$ is an isotropic series, if $\frac { a _ { 11 } } { a _ { 10 ...

The series $\left\{ a _ { n } \right\}$ is an isotropic series, if $\frac { a _ { 11 } } { a _ { 10 } } < - 1$, and the sum of the first $n$ terms of $S _ { n }$ has a maximum, then when $S _ { n }$ obtains When $S _ { n }$ gets the smallest positive value, $n =$

  • A. A. 11
  • B. B. 17
  • C. C. 19
  • D. D. 21

Answer: C

Solution: Test analysis : $\because \mathrm { Sn }$ has maximum value, $\therefore \mathrm { d } < 0$ then $\mathrm { a } _ { 10 } > \mathrm { a } _ { 11 }$, and $\frac { a _ { 11 } < - 1 } { a _ { 10 } } , \therefore \mathrm { a } _ { 11 } < 0 <$ $\mathrm { a } _ { 10 } \therefore \mathrm { a } _ { 10 } + \mathrm { a } _ { 11 } < 0$, $\mathrm { a } _ { 10 } \therefore \mathrm { a } _ { 10 } + \mathrm { a } _ { 11 } < 0$, $\therefore S _ { 20 } = 10 \left( a _ { 1 } + a _ { 20 } \right) = 10 \left( a _ { 10 } + a _ { 11 } \right) < 0 \quad , S _ { 19 } = 19 a _ { 10 } > 0 \quad$ $\therefore S _ { 20 } = 10 \left( a _ { 1 } + a _ { 20 } \right) = 10 \left( a _ { 10 } + a _ { 11 } \right) < 0 \quad , S _ { 19 } = 19 a _ { 10 } > 0 \quad$ also $a _ { 1 } > a _ { 2 } > \cdots > a _ { 10 } > 0 > a _ { 11 } > a _ { 12 }$ $\therefore \quad S _ { 10 } > S _ { 9 } > \cdots > S _ { 2 } > S _ { 1 } > 0 \quad , \quad S _ { 10 } > S _ { 11 } > \cdots > S _ { 19 } > 0 > S _ { 20 } > S _ { 21 }$ $S _ { 19 } - S _ { 1 } = a _ { 2 } + a _ { 3 } + \cdots + a _ { 19 } = 9 \left( a _ { 10 } + a _ { 11 } \right) < 0 \quad \therefore S _ { 19 }$ is the smallest positive value. Point : Properties of Equal Difference Series

Question 22: 23. If the sum of the first $n$ terms of the series $\left\{ a _ { n } \right\}$ is $S _ { n }$ and ...

23. If the sum of the first $n$ terms of the series $\left\{ a _ { n } \right\}$ is $S _ { n }$ and satisfies $S _ { n } = \frac { 1 } { 2 } \left( a _ { n } + n - 1 \right)$, then the sum of the first 81 terms of the series $\left\{ n a _ { n } \right\}$ is ( )

  • A. A. 1640
  • B. B. 1660
  • C. C. 1680
  • D. D. 1700

Answer: A

Solution: From $S _ { n } = \frac { 1 } { 2 } \left( a _ { n } + n - 1 \right)$ , the we have $a _ { n + 1 } = S _ { n + 1 } - S _ { n } = \frac { 1 } { 2 } \left( a _ { n + 1 } + n \right) - \frac { 1 } { 2 } \left( a _ { n } + n - 1 \right) = \frac { 1 } { 2 } a _ { n + 1 } - \frac { 1 } { 2 } a _ { n } + \frac { 1 } { 2 }$ and we have $a _ { n } + a _ { n + 1 } = 1$. And from $a _ { 1 } = S _ { 1 } = \frac { 1 } { 2 } a _ { 1 }$, we have $a _ { 1 } = 0$, and we have $a _ { n } = \left\{ \begin{array} { l } 0 , n \text { 为奇数 } \\ 1 , n \text { 为偶数 } \end{array} \right.$. The sum of the first 81 terms of the series $\left\{ n a _ { n } \right\}$ is $2 + 4 + 6 + \cdots + 80 = \frac { 40 \times ( 2 + 80 ) } { 2 } = 1640$.

Question 23: 24. In the isomorphic series $\left\{ a _ { n } \right\}$ with tolerance ${ } _ { d }$ and $S _ { 10...

24. In the isomorphic series $\left\{ a _ { n } \right\}$ with tolerance ${ } _ { d }$ and $S _ { 10 } = 4 S _ { 5 }$, then $\frac { a _ { 1 } } { d }$ is equal to

  • A. A. $\frac { 1 } { 4 }$
  • B. B. 8
  • C. C. $\frac { 1 } { 2 }$
  • D. D. 4

Answer: C

Solution: According to the question, there is $10 a _ { 1 } + 45 d = 4 \left( 5 a _ { 1 } + 10 d \right)$, which is $10 a _ { 1 } = 5 d$, so $\frac { a _ { 1 } } { d } = \frac { 1 } { 2 }$. So choose C. [点睛]本小题主要考查等差数列的前 $n$ 项和公式,利用等差数列前 $n$ 项和公式求比值. It is a basic question, and the result can be calculated by directly substituting the formula.

Question 24: 25. If $S _ { n }$ is known to be the sum of the first $n$ terms of the isomorphic series $\left\{ a...

25. If $S _ { n }$ is known to be the sum of the first $n$ terms of the isomorphic series $\left\{ a _ { n } \right\}$ and $S _ { 3 } = 2 a _ { 1 }$, then the following conclusion is false

  • A. A. $a _ { 4 } = 0$
  • B. B. $S _ { 4 } = S _ { 3 }$
  • C. C. $S _ { 7 } = 0$
  • D. D. $\left\{ a _ { n } \right\}$ is a decreasing number series

Answer: D

Solution: Test analysis: Let the tolerance of the equal difference series $\left\{ a _ { n } \right\}$ be $d$. From $S _ { 3 } = 2 a _ { 1 }$, we can obtain: $a _ { 1 } + a _ { 2 } + a _ { 3 } = 3 a _ { 1 } + 3 d = 2 a _ { 1 }$, which can be $a _ { 1 } = - 3 d$. Using the generalization and summation formulas, you can determine the correctness of $A$, B, and C. Since it is impossible to determine the $d$ positive or negative, it is impossible to determine the monotonicity of the isomorphic series $\left\{ a _ { n } \right\}$, so you can determine the correctness or incorrectness of D. Explanation: Let the tolerance of the equal difference series $\left\{ a _ { n } \right\}$ be $d$. From $S _ { 3 } = 2 a _ { 1 }$, we have: $a _ { 1 } + a _ { 2 } + a _ { 3 } = 3 a _ { 1 } + 3 d = 2 a _ { 1 }$, which gives $a _ { 1 } = - 3 d$. Then $\mathrm { a } _ { 4 } = - 3 \mathrm {~d} + 3 \mathrm {~d} = 0 , \mathrm {~S} _ { 4 } = \mathrm { S } _ { 3 } , \mathrm {~S} _ { 7 } = \frac { 7 \left( a _ { 1 } + a _ { 7 } \right) } { 2 } = 7 \mathrm { a } _ { 4 } = 0$ , so A, B, and C are correct. Since it is not possible to determine the positivity or negativity of d, it is not possible to determine the monotonicity of the equal-difference series $\left\{ \mathrm { a } _ { \mathrm { n } } \right\}$, so D is incorrect.

Question 25: 26. Read the block diagram on the right. If the input $n$ is 100, then the values of the output vari...

26. Read the block diagram on the right. If the input $n$ is 100, then the values of the output variables $S$ and $T$ are, in order, ( ) ![](/images/questions/sequence/image-002.jpg)

  • A. A. 2450, 2500
  • B. B. 2550, 2450
  • C. C. 2500, 2550
  • D. D. 2550, 2500

Answer: D

Solution: Simulate the execution of the block diagram to obtain $n = 100 , S = 0 , T = 0$; The condition $_ { n < 2 } , S = 100 , n = 99 , T = 99 , n = 98$ is not satisfied; The condition $_ { n < 2 } , S = 100 + 98 , n = 97 , T = 99 + 97 , n = 96$ is not satisfied; The condition $_ { n < 2 } , S = 100 + 98 + 96 , n = 95 , T = 99 + 97 + 95 , n = 94$ is not satisfied; L The condition $_ { n < 2 } , S = 100 + 98 + \cdots + 4 , n = 3 , T = 99 + 97 + \cdots + 3 , n = 2$ is not satisfied; The condition $_ { n < 2 } , S = 100 + 98 + \cdots + 4 + 2 , n = 1 , T = 99 + 97 + \cdots + 3 + 1 , n = 0$ is not satisfied; Satisfy condition ${ } _ { n < 2 }$, exit the loop, and output the value of $S , T$. So $S = 100 + 98 + 96 + \cdots + 2 = 2550$ , $S = 100 + 98 + 96 + \cdots + 2 = 2550$ $T = 99 + 97 + \cdots + 1 = 2500$.

Question 26: 27. Given the following propositions, the number of correct propositions is ( ) (1) There exist two ...

27. Given the following propositions, the number of correct propositions is ( ) (1) There exist two unequal real numbers $\alpha , \beta$ such that the equation $\sin ( \alpha + \beta ) = \sin \alpha + \sin \beta$ holds; (2) If the series $\left\{ a _ { n } \right\}$ is isometric and $m + n = s + t , m , n , s , t \in N ^ { * }$, then $a _ { m } + a _ { n } = a _ { s } + a _ { t }$; (3) If $S _ { n }$ is the sum of the first $n$ terms of the isoperimetric series $\left\{ a _ { n } \right\}$ and $S _ { n } = 3 \cdot 2 ^ { n } + A$, then $A = - 3$; (4) It is known that the sides opposite the three interior angles $A , B , C$ of $a , b , c$ are $a , b , c$ , and if $a ^ { 2 } + b ^ { 2 } > c ^ { 2 }$ , then $V A B C -$ is definitely an an acute triangle; ( )

  • A. A. 1
  • B. B. 2
  • C. C. 3
  • D. D. 4

Answer: C

Solution: For (1), $\alpha = 0 , \beta = \pi$ is correct if $\sin ( \alpha + \beta ) = \sin \alpha + \sin \beta$; For (2), when the series $\left\{ a _ { n } \right\}$ is an isotropic series, let the tolerance be $d$, if $m + n = s + t$, then $m + n = s + t$ then $a _ { m } + a _ { n } = a _ { 1 } + ( m - 1 ) d + a _ { 1 } + ( n - 1 ) d = 2 a _ { 1 } + ( m + n - 2 ) d$, $a _ { m } + a _ { n } = a _ { 1 } + ( m - 1 ) d + a _ { 1 } + ( n - 1 ) d = 2 a _ { 1 } + ( m + n - 2 ) d$, $a _ { m } + a _ { n } = a _ { 1 } + ( m - 1 ) d + a _ { 1 } + ( n - 1 ) d = 2 a _ { 1 } + ( m + n - 2 ) d$ $a _ { s } + a _ { t } = a _ { 1 } + ( s - 1 ) d + a _ { 1 } + ( t - 1 ) d = 2 a _ { 1 } + ( s + t - 2 ) d$ Therefore $a _ { m } + a _ { n } = a _ { s } + a _ { t }$, is correct; For (3), $a _ { n } = S _ { n } - S _ { n - 1 } = \left( 302 ^ { n } + A \right) - \left( 302 ^ { n - 1 } + A \right) = 302 ^ { n - 1 }$ and $S _ { 1 } = 6 + A = a _ { 1 } = 3$, so $A = - 3$, is correct; For (4), take ${ } ^ { a = 5 , b = 4 , c = 3 }$, it is easy to know that $a ^ { 2 } + b ^ { 2 } > c ^ { 2 }$ , and the triangle is a right triangle, so it is wrong.

Question 27: 28. It is known that the isoperimetric series $\left\{ a _ { n } \right\}$ satisfies $a _ { 1 } = 1 ...

28. It is known that the isoperimetric series $\left\{ a _ { n } \right\}$ satisfies $a _ { 1 } = 1 , a _ { 5 } = 4$, then $a _ { 2 } a _ { 3 } a _ { 4 } = ( )$

  • A. A. - 8
  • B. B. - 16
  • C. C. 8
  • D. D. 16

Answer: C

Solution: By the isoperimetric series property, we can get $a _ { 3 } ^ { 2 } = a _ { 2 } a _ { 4 } = a _ { 1 } a _ { 5 } = 4$ , ${ } ^ { a _ { 1 } = 1 }$ Also ${ } ^ { a _ { 1 } = 1 }$ , so $a _ { 3 } = a _ { 1 } q ^ { 2 } > 0$ , so $a _ { 3 } = 2$ , so $a _ { 2 } a _ { 3 } a _ { 4 } = 4 \times 2 = 8$ . so $a _ { 2 } a _ { 3 } a _ { 4 } = 4 \times 2 = 8$.

Question 28: It is known that the series $\left\{ a _ { n } \right\}$ satisfies $a _ { n } = 3 \times 2 ^ { n - 1...

It is known that the series $\left\{ a _ { n } \right\}$ satisfies $a _ { n } = 3 \times 2 ^ { n - 1 } , n \in N ^ { * }$, which is now arranged in a serpentine array according to the following pattern (row $i$). INLINE_FORMULA_3]] and $i \in N ^ { * }$), and the ${ } ^ { j }$ number in the $i$ row from the left is ${ } ^ { ~ } { } _ { ( i , j ) } \left( i , j \in N ^ { * } \right.$ and $\left. { } ^ { j \leq i } \right)$ and $a _ { ( 21,21 ) } =$ is $a _ { n } = 3 \times 2 ^ { n - 1 } , n \in N ^ { * }$, and $a _ { n } = 3 \times 2 ^ { n - 1 } , n \in N ^ { * }$ is $i$. FORMULA_8]], then $a _ { ( 21,21 ) } =$ $a _ { 1 }$ $a _ { 2 } \quad a _ { 3 }$ $\begin{array} { l l l } a _ { 6 } & a _ { 5 } & a _ { 4 } \end{array}$ $\begin{array} { l l l l } a _ { 7 } & a _ { 8 } & a _ { 9 } & a _ { 10 } \end{array}$ $\begin{array} { l l l l l } a _ { 15 } & a _ { 14 } & a _ { 13 } & a _ { 12 } & a _ { 11 } \end{array}$ $\_\_\_\_$

  • A. A. $3 \times 2 ^ { 209 }$
  • B. B. $3 \times 2 ^ { 230 }$
  • C. C. $3 \times 2 ^ { 211 }$
  • D. D. $3 \times 2 ^ { 212 }$

Answer: B

Solution: SOLUTION: From the question, there are ${ } _ { i }$ numbers in line ${ } _ { i }$, then there are $\frac { ( 1 + 21 ) \times 21 } { 2 } = 231$ numbers in the first 21 lines, and $a _ { ( 21,21 ) }$ corresponds to the last number of line 21, i.e., $a _ { 231 }$, i.e., $a _ { ( 21,21 ) } = a _ { 231 } = 3 \times 2 ^ { 230 }$. FORMULA_4]], which is $a _ { ( 21,21 ) } = a _ { 231 } = 3 \times 2 ^ { 230 }$.

Question 29: 30. The series $\left\{ a _ { n } \right\}$ is known to be an isoperimetric series, if $a _ { 5 } - ...

30. The series $\left\{ a _ { n } \right\}$ is known to be an isoperimetric series, if $a _ { 5 } - a _ { 3 } = 12 , a _ { 6 } - a _ { 4 } = 24$, then $a _ { 2024 } =$

  • A. A. $2 ^ { 2023 } - 1$
  • B. B. $2 ^ { 2023 }$
  • C. C. $2 ^ { 2024 } - 1$
  • D. D. $2 ^ { 2024 }$

Answer: B

Solution: Let the common ratio of the isoperimetric series be $q$ since $a _ { 5 } - a _ { 3 } = 12 , a _ { 6 } - a _ { 4 } = 24$ , and So from $a _ { 6 } - a _ { 4 } = q \left( a _ { 5 } - a _ { 3 } \right)$ , we have $24 = 12 q$ , so $q = 2$ , . and $a _ { 1 } q ^ { 4 } - a _ { 1 } q ^ { 2 } = 12$, which is $a _ { 1 } \times 2 ^ { 4 } - a _ { 1 } \times 2 ^ { 2 } = 12$, so ${ } ^ { a _ { 1 } = 1 }$ , and ${ } _ { 2024 } = 1 \times 2 ^ { 2024 - 1 } = 2 ^ { 2023 }$, which is $a _ { 1 } \times 2 ^ { 4 } - a _ { 1 } \times 2 ^ { 2 } = 12$. so ${ } _ { 2024 } = 1 \times 2 ^ { 2024 - 1 } = 2 ^ { 2023 }$.

Question 30: 31. It is known that the sum of the first $n$ terms of the isomorphic series $\left\{ a _ { n } \rig...

31. It is known that the sum of the first $n$ terms of the isomorphic series $\left\{ a _ { n } \right\}$ is $S _ { n }$ and $a _ { 1 } + a _ { 3 } + a _ { 8 } = 6$ is $S _ { 7 } =$.

  • A. A. 28
  • B. B. 21
  • C. C. 16
  • D. D. 14

Answer: D

Solution: Because $a _ { 1 } + a _ { 3 } + a _ { 8 } = 3 a _ { 1 } + 9 d = 3 a _ { 4 } = 6$, therefore $a _ { 4 } = 2 , S _ { 7 } = \frac { 7 \left( a _ { 1 } + a _ { 7 } \right) } { 2 } = 7 a _ { 4 } = 14$.

Question 31: 32. If $\left\{ a _ { n } \right\}$ is known to be an isoperimetric series, and the middle term of t...

32. If $\left\{ a _ { n } \right\}$ is known to be an isoperimetric series, and the middle term of the equality between $a _ { 2 } \cdot a _ { 3 } = 2 a _ { 1 } , a _ { 4 }$ and $2 a _ { 7 }$ is $\frac { 5 } { 4 }$, then $a _ { 5 } =$ is an isoperimetric series.

  • A. A. 1
  • B. B. 2
  • C. C. 31
  • D. D. $\frac { 1 } { 2 }$

Answer: A

Solution: From ${ } ^ { a _ { 2 } } \cdot a _ { 3 } = 2 a _ { 1 }$ to ${ } ^ { a _ { 1 } q ^ { 3 } = a _ { 4 } = 2 }$. and $a _ { 4 } + 2 a _ { 7 } = 2 + 2 a _ { 7 } = \frac { 5 } { 2 }$ to get $2 a _ { 1 } q ^ { 6 } = \frac { 1 } { 2 }$ , (2) From (1) (2), we have $a _ { 1 } = 16 , q = \frac { 1 } { 2 } , \therefore a _ { 5 } = 16 \times \left( \frac { 1 } { 2 } \right) ^ { 4 } = 1$.

Question 32: 34. If all the terms of the isoperimetric series $\left\{ a _ { n } \right\}$ are positive, and $a _...

34. If all the terms of the isoperimetric series $\left\{ a _ { n } \right\}$ are positive, and $a _ { 5 } a _ { 6 } + a _ { 4 } a _ { 7 } = 6$, then $\log _ { 3 } \left( a _ { 1 } a _ { 2 } \cdots a _ { 10 } \right) =$

  • A. A. 1
  • B. B. 5
  • C. C. 15
  • D. D. 30

Answer: B

Solution: Since the series $\left\{ a _ { n } \right\}$ is an isoperimetric series, $a _ { 5 } a _ { 6 } + a _ { 4 } a _ { 7 } = 2 a _ { 5 } a _ { 6 } = 6 , a _ { 5 } a _ { 6 } = 3$, therefore $a _ { 1 } a _ { 2 } \cdots a _ { 10 } = \left( a _ { 5 } a _ { 6 } \right) ^ { 5 } = 3 ^ { 5 }$, therefore $\log _ { 3 } \left( a _ { 1 } a _ { 2 } \cdots a _ { 10 } \right) = \log _ { 3 } 3 ^ { 5 } = 5$.

Question 33: 35. The sum of the first $n$ terms of the series $\left\{ a _ { n } \right\}$ is $S _ { n } , a _ { ...

35. The sum of the first $n$ terms of the series $\left\{ a _ { n } \right\}$ is $S _ { n } , a _ { 1 } = 1 , a _ { n } = \left\{ \begin{array} { l } a _ { n - 1 } + 1 , n = 2 k \\ 2 a _ { n - 1 } + 1 , n = 2 k + 1 \end{array} \left( k \in \mathrm {~N} ^ { * } \right) \right.$. The following options are correct Ratio series

  • A. A. $a _ { 6 } = 16$
  • B. B. The series $\left\{ a _ { 2 k } + 3 \right\} \left( k \in \mathrm {~N} ^ { * } \right)$ is an equal series with 2 as the common ratio.
  • C. C. For any $k \in \mathrm {~N} ^ { * } , a _ { 2 k } = 2 ^ { k + 1 } - 3$
  • D. D. The smallest positive integer of $S _ { n } > 1000$ $n$ has a value of 15.

Answer: D

Solution: According to the question, $a _ { 2 k } - a _ { 2 k - 1 } = 1 , a _ { 2 k + 1 } - 2 a _ { 2 k } = 1$ , from ${ } ^ { a _ { 1 } = 1 } , ~ a _ { 2 } - a _ { 1 } = 1$ , we get $a _ { 2 } = a _ { 1 } + 1 = 2$ , and $a _ { 2 k + 2 } - a _ { 2 k + 1 } = 1$ , and $a _ { 2 k + 2 } - 2 a _ { 2 k } = 2$ , that is $a _ { 2 k + 2 } + 2 = 2 \left( a _ { 2 k } + 2 \right)$ . Also $a _ { 2 k + 2 } - a _ { 2 k + 1 } = 1$, then $a _ { 2 k + 2 } - 2 a _ { 2 k } = 2$, that is, $a _ { 2 k + 2 } + 2 = 2 \left( a _ { 2 k } + 2 \right)$, and $a _ { 2 } + 2 = 4 \neq 0$, so the series $\left\{ a _ { 2 k } + 2 \right\}$ is equiprobable, and $a _ { 2 k } + 2 = 4 \times 2 ^ { k - 1 }$ FORMULA_8]], i.e., $a _ { 2 k } = 2 ^ { k + 1 } - 2 , a _ { 6 } = 16 - 2 = 14$, AC is wrong; For B, $a _ { 2 k } + 3 = 2 ^ { k + 1 } + 1 , \frac { a _ { 2 k + 2 } + 3 } { a _ { 2 k } + 3 } = \frac { 2 ^ { k + 2 } + 1 } { 2 ^ { k + 1 } + 1 }$ is not a constant and the series $\left\{ a _ { 2 k } + 3 \right\}$ is not an isoperimetric series, B is wrong; For D, $a _ { 2 k - 1 } = a _ { 2 k } - 1 = 2 ^ { k + 1 } - 3$, the terms of the series $\left\{ a _ { n } \right\}$ are all positive, so the series $\left\{ S _ { n } \right\}$ is an increasing series. $S _ { 14 } = a _ { 1 } + a _ { 2 } + \cdots + a _ { 14 } = a _ { 1 } + \left( a _ { 1 } + 1 \right) + a _ { 3 } + \left( a _ { 3 } + 1 \right) + \cdots + a _ { 13 } + \left( a _ { 13 } + 1 \right)$ $= 2 \left( a _ { 1 } + a _ { 3 } + a _ { 5 } + a _ { 7 } + a _ { 9 } + a _ { 11 } + a _ { 13 } \right) + 7 = 2 \times \left( 2 ^ { 2 } - 3 + 2 ^ { 3 } - 3 + \cdots + 2 ^ { 8 } - 3 \right) + 7 = 981$ $S _ { 15 } = S _ { 14 } + a _ { 15 } = 981 + 509 = 1490 > 1000$, so the smallest positive integer $n$ of ${ } ^ { S _ { n } > 1000 }$ has the value 15.

Question 34: 36. The positive isometric series $\left\{ a _ { n } \right\}$ is known to satisfy $2 a _ { 4 } + a ...

36. The positive isometric series $\left\{ a _ { n } \right\}$ is known to satisfy $2 a _ { 4 } + a _ { 3 } = a _ { 2 }$ , if there are two terms in the series $\left\{ a _ { n } \right\}$ and the equivariant middle term of $a _ { m } , a _ { n }$ is $\frac { a _ { 1 } } { 4 }$, then The minimum value of $\frac { 4 } { m } + \frac { 1 } { n }$ is ( )

  • A. A. $\frac { 3 } { 2 }$
  • B. B. $\frac { 5 } { 3 }$
  • C. C. 2
  • D. D. $\frac { 13 } { 6 }$

Answer: A

Solution: Since $2 a _ { 4 } + a _ { 3 } = a _ { 2 }$, let the first term of the isoperimetric series $\left\{ a _ { n } \right\}$ be ${ } ^ { a _ { 1 } }$ and the common ratio be $q$, then $2 a _ { 1 } \cdot q ^ { 3 } + a _ { 1 } \cdot q ^ { 2 } = a _ { 1 } \cdot q$, so $2 q ^ { 2 } + q = 1$, which solves : $q = - 1$ (rounded) or $q = \frac { 1 } { 2 }$, and because $\frac { a _ { 1 } } { 4 }$ is the first term of $a _ { m }$, the common ratio is $\left\{ a _ { n } \right\}$. FORMULA_9]] $a _ { n }$ is the equivocal middle term of $\left( \frac { a _ { 1 } } { 4 } \right) ^ { 2 } = a _ { m } \cdot a _ { n }$, then $a _ { 1 } ^ { 2 } = 16 a _ { 1 } \cdot q ^ { m - 1 } \cdot a _ { 1 } \cdot q ^ { n - 1 }$, so $\left( \frac { 1 } { 2 } \right) ^ { m + n - 2 } = \frac { 1 } { 16 } = \left( \frac { 1 } { 2 } \right) ^ { 4 }$, and i.e. $m + n = 6$, so $\frac { 4 } { m } + \frac { 1 } { n } = \frac { 1 } { 6 } ( m + n ) \times \left( \frac { 4 } { m } + \frac { 1 } { n } \right) = \frac { 1 } { 6 } \left( 4 + \frac { m } { n } + \frac { 4 n } { m } + 1 \right) = \frac { 1 } { 6 } \left( 5 + \frac { m } { n } + \frac { 4 n } { m } \right)$ $\geq \frac { 1 } { 6 } \left( 5 + 2 \sqrt { \frac { m } { n } \cdot \frac { 4 n } { m } } \right) = \frac { 3 } { 2 }$, which is equal if and only if $\frac { m } { n } = \frac { 4 n } { m }$, i.e. $n = 2 , m = 4$.

Question 35: 37. It is known that the sum of the first $n$ terms of the isomorphic series $\left\{ a _ { n } \rig...

37. It is known that the sum of the first $n$ terms of the isomorphic series $\left\{ a _ { n } \right\}$ is $n$ and $S _ { n }$ and $S _ { 25 } = 100$ is $a _ { 12 } + a _ { 14 } =$.

  • A. A. 16
  • B. B. 8
  • C. C. 4
  • D. D. 2

Answer: B

Solution: Since $S _ { 25 } = \frac { 25 } { 2 } \left( a _ { 1 } + a _ { 25 } \right) = 25 a _ { 13 } = 100$ $\therefore a _ { 13 } = 4$ $\therefore a _ { 12 } + a _ { 14 } = 2 a _ { 13 } = 8$, so choose B. [点睛]本题主要考查等差数列的性质以及前 $n$ 项和公式的应用,属于中档题. When solving problems about the equal difference series, pay attention to the application of the properties of the equal difference series $a _ { p } + a _ { q } = a _ { m } + a _ { n } = 2 a _ { r } \quad ( p + q = m + n = 2 r )$ and the relationship between the sum of the first $n$ terms.

Question 36: In the isoperimetric series $\left\{ a _ { n } \right\}$, $a _ { 1 } + a _ { 2 } + a _ { 3 } + a _ {...

In the isoperimetric series $\left\{ a _ { n } \right\}$, $a _ { 1 } + a _ { 2 } + a _ { 3 } + a _ { 4 } = 20 , a _ { 5 } + a _ { 6 } + a _ { 7 } + a _ { 8 } = 10$ is known, and the sum of the first 16 terms of the series $\left\{ a _ { n } \right\}$ is $S _ { 16 }$.

  • A. A. 20
  • B. B. $\frac { 75 } { 2 }$
  • C. C. $\frac { 125 } { 2 }$
  • D. D. $- \frac { 75 } { 2 }$

Answer: B

Solution: 试题分析:由题意得,$S _ { 4 } = 20 , S _ { 8 } - S _ { 4 } = 10$, 則 $\frac { S _ { 8 } - S _ { 4 } } { S _ { 4 } } = \frac { 1 } { 2 }$ , according to the nature of the isoperimetric series It can be shown that $S _ { 4 } , S _ { 8 } - S _ { 4 } , S _ { 12 } - S _ { 8 } , S _ { 16 } - S _ { 12 }$ forms an isometric series with common ratio $\frac { 1 } { 2 }$. $S _ { 4 } = 20 , S _ { 8 } - S _ { 4 } = 10 , S _ { 12 } - S _ { 8 } = 5 , S _ { 16 } - S _ { 12 } = \frac { 5 } { 2 }$ and $S _ { 8 } = 30 , S _ { 12 } = 35 , S _ { 16 } = \frac { 75 } { 2 }$. Points: the properties of the isoperimetric series.

Question 37: 39. If the series $\left\{ a _ { n } \right\}$ is an isometric series, $\left\{ b _ { n } \right\}$ ...

39. If the series $\left\{ a _ { n } \right\}$ is an isometric series, $\left\{ b _ { n } \right\}$ is an isoperimetric series, and satisfies: $a _ { 1 } + a _ { 2019 } = \pi , b _ { 1 } \cdot b _ { 2019 } = 2$ , the function $f ( x ) = \sin x$, then $f \left( \frac { a _ { 1009 } + a _ { 1011 } } { 1 + b _ { 1009 } b _ { 1011 } } \right) =$

  • A. A. $- \frac { \sqrt { 3 } } { 2 }$
  • B. B. $\frac { 1 } { 2 }$
  • C. C. $\frac { \sqrt { 3 } } { 2 }$
  • D. D. $- \frac { 1 } { 2 }$

Answer: C

Solution: By the property of the isomorphic series we get $a _ { 1 } + a _ { 2019 } = a _ { 1009 } + a _ { 1011 } = \pi$ and by the property of the isoperimetric series we have $b _ { 1 } \cdot b _ { 2019 } = b _ { 1009 } \cdot b _ { 1011 } = 2 ; f \left( \frac { a _ { 1009 } + a _ { 1011 } } { 1 + b _ { 1009 } b _ { 1011 } } \right) = f \left( \frac { \pi } { 3 } \right) = \frac { \sqrt { 3 } } { 2 }$. Therefore, the question is C. [EYE POINT] This question examines the application of the nature of isometric and isometric series, is the basic calculation problems, for isometric and isometric series of small problems, commonly used methods, one is to the basic quantities of the first term and the common ratio or tolerance, and the second is to observe the foot code relationship between the items, that is, the use of the basic nature of the series.

Question 38: 40. The series $\left\{ a _ { n } \right\}$ satisfies $a _ { n + 1 } = \left( 2 \left| \sin \frac { ...

40. The series $\left\{ a _ { n } \right\}$ satisfies $a _ { n + 1 } = \left( 2 \left| \sin \frac { n \pi } { 2 } \right| - 1 \right) a _ { n } + n , n \in \mathrm {~N} ^ { * }$, then the sum of the first 20 terms of the series $\left\{ a _ { n } \right\}$ is High School Mathematics Assignment, October 29, 2025

  • A. A. 100
  • B. B. 110
  • C. C. 160
  • D. D. 200

Answer: B

Solution: SOLUTION: From $a _ { n + 1 } = \left( 2 \left| \sin \frac { n \pi } { 2 } \right| - 1 \right) a _ { n } + n$. we get: $a _ { 1 } = a _ { 1 } , a _ { 2 } = a _ { 1 } + 1 , a _ { 3 } = - a _ { 2 } + 2 = - a _ { 1 } + 1 , a _ { 4 } = a _ { 3 } + 3 = - a _ { 1 } + 4$ $\therefore ^ { a _ { 1 } + a _ { 2 } + a _ { 3 } + a _ { 4 } = 6 }$. Similarly, $a _ { 5 } + a _ { 6 } + a _ { 7 } + a _ { 8 } = 14 , a _ { 9 } + a _ { 10 } + a _ { 11 } + a _ { 12 } = 22$ is obtained from $a _ { 1 } = a _ { 1 } , a _ { 2 } = a _ { 1 } + 1 , a _ { 3 } = - a _ { 2 } + 2 = - a _ { 1 } + 1 , a _ { 4 } = a _ { 3 } + 3 = - a _ { 1 } + 4$ and $\therefore ^ { a _ { 1 } + a _ { 2 } + a _ { 3 } + a _ { 4 } = 6 }$. The first 20 terms of the series $\therefore$ satisfy $\left\{ a _ { n } \right\}$ that $S _ { 4 } , S _ { 8 } - S _ { 4 } , S _ { 12 } - S _ { 8 } , \cdots$ is an isotropic series with 6 as the first term and 8 as the tolerance, then the sum of the first 20 terms of the series $\left\{ a _ { n } \right\}$ is $S = 5 \times 6 + \frac { 5 \times 4 } { 2 } \times 8 = 110$. FORMULA_8]].
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Sequence

数列

38 Practice Questions

Practice with Chinese questions to prepare for the CSCA exam. You can toggle translations while practicing.

Topic Overview

A series is a sequence of numbers in a certain order in mathematics, mainly including equal difference series and equal ratio series. In the CSCA exam, series questions often involve the general formula, the calculation of the sum of the first n terms, and the application of series properties in geometric and algebraic problems. Mastering the basic formulas and applying them flexibly is the key to solving the questions.

Questions:38

Key Points

  • 1The generalization and summation formulas of the equivariant series
  • 2The general and summation formulas of the isoperimetric series
  • 3Calculation of Equivalent and Equivalent Median Terms
  • 4Integration of series with other knowledge points (e.g. conic curves)

Study Tips

It is recommended to familiarize yourself with the core formulas of the equal-difference and equal-ratio series, and to master the property of the middle term and the solution of cross-point synthesis problems through practice problems.

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