answer:To find all arithmetic progressions for which the sums of the first ( n_1 ) and ( n_2 ) terms are ( n_1^2 ) and ( n_2^2 ) respectively, we will start by using the formula for the sum of the first ( n ) terms of an arithmetic progression. 1. **Sum Formula**: The sum of the first ( n ) terms of an arithmetic progression can be given by: [ S_n = frac{n}{2} left(2a_1 + (n-1)d right) ] where ( a_1 ) is the first term and ( d ) is the common difference. 2. **Given Conditions**: According to the problem, we have: [ S_{n_1} = n_1^2 quad text{and} quad S_{n_2} = n_2^2 ] 3. **Applying the Sum Formula**: Substitute the sum formula into the given conditions: [ frac{n_1}{2} left(2a_1 + (n_1-1)d right) = n_1^2 ] [ frac{n_2}{2} left(2a_1 + (n_2-1)d right) = n_2^2 ] 4. **Simplification**: Simplify these equations by multiplying both sides by 2 and then dividing by ( n_1 ) and ( n_2 ) respectively: [ 2a_1 + (n_1-1)d = 2n_1 ] [ 2a_1 + (n_2-1)d = 2n_2 ] 5. **Isolate (a_1)**: Rearrange both equations to express in terms of ( a_1 ): [ 2a_1 = 2n_1 - (n_1-1)d ] [ 2a_1 = 2n_2 - (n_2-1)d ] 6. **Equate the Equations**: Since ( 2a_1 ) is the same in both equations: [ 2n_1 - (n_1-1)d = 2n_2 - (n_2-1)d ] 7. **Solving for (d)**: Rearrange to isolate terms involving (d): [ 2n_1 - 2n_2 = (n_1-1)d - (n_2-1)d ] [ 2(n_1-n_2) = (n_1 - n_2)d ] Given that ( n_1 neq n_2 ), factor out ( n_1 - n_2 ): [ 2 = d ] 8. **Substitute ( d ) Back**: Use ( d = 2 ) in one of the simplified equations to find ( a_1 ): [ 2a_1 = 2n_1 - (n_1-1) cdot 2 ] [ 2a_1 = 2n_1 - 2n_1 + 2 ] [ 2a_1 = 2 ] [ a_1 = 1 ] # Conclusion Therefore, the arithmetic progression has the first term ( a_1 = 1 ) and the common difference ( d = 2 ). Thus, the arithmetic progression is: [ 1, 3, 5, 7, 9, ldots ] [ boxed{1, 3, 5, 7, 9, ldots} ]

answer:Let the price of a tulip be x yuan and the price of a lilac be y yuan. According to the problem, we have: 1. 4x + 5y < 22 2. 6x + 3y > 24 We need to compare 2x and 3y. From the first inequality, we can derive that 2x + 2.5y < 11. This means that if we buy 2 tulips and 2.5 lilacs, the cost is less than 11 yuan. However, since we cannot actually buy half a lilac, this information suggests that buying 2 tulips is relatively cheaper compared to buying 3 lilacs, given the constraints. From the second inequality, we can derive that 2x + y > 8. This means that the combined price of 2 tulips and 1 lilac is more than 8 yuan. This further implies that tulips are relatively more expensive, as the combination of fewer tulips and lilacs still results in a higher total cost. Considering the given inequalities and the deductions made, it is evident that the cost of buying 2 tulips is higher than the cost of buying 3 lilacs under the constraints provided by the problem. Therefore, the correct answer is boxed{text{A: The former is more expensive}}.

answer:Since the function f(x) satisfies that for any x_{1} < x_{2}, f(x_{1}) < f(x_{2}) holds, Therefore, the function f(x) is an increasing function on its domain, Then it satisfies begin{cases} 2-a > 0 a > 1 2-a+1leqslant aend{cases}, which is begin{cases} a < 2 a > 1 ageqslant dfrac {3}{2}end{cases}, we get dfrac {3}{2}leqslant a < 2, Hence, the answer is: boxed{dfrac {3}{2}leqslant a < 2}. By judging that the function is increasing based on the condition, and solving it by establishing inequality relationships according to the definition of function monotonicity. This question mainly examines the application of function monotonicity. Establishing inequality relationships based on the properties of monotonicity of piecewise functions is the key to solving this question.

answer:The smallest number of candies must be the least common multiple of the numbers 2, 3, 4, 5, 6, 7, and 9. Prime factorizing: begin{align*} 2 & = 2 3 & = 3 4 & = 2^2 5 & = 5 6 & = 2 cdot 3 7 & = 7 9 & = 3^2. end{align*} To find the LCM, we take the highest powers of all prime factors appearing in these factorizations: - 2^2 from 4 - 3^2 from 9 - 5^1 from 5 - 7^1 from 7 Hence, the LCM is 2^2 cdot 3^2 cdot 5 cdot 7 = 4 cdot 9 cdot 5 cdot 7 = 36 cdot 35 = 1260. Therefore, the least number of candies is boxed{1260}.