Introduction to APs and common difference.

The fourth term is 7+3d, and it is given to be 34. Set up the equation 7+3d=34. Subtracting 7 gives 3d=27, so d=9. The second term is 7+d, hence it equals 7+9=16. Therefore option D is correct. The full sequence is 7, 16, 25, 34, and each pair of consecutive terms differs by 9, confirming the AP structure. A common error is to divide the total change 34−7 by 4; however, four terms create only three equal gaps, so the correct division is by 3. The governing concept is the standard AP form a, a+d, a+2d, a+3d, where the coefficient of d indicates the number of gaps from the first term.

Four consecutive terms of an arithmetic progression contain three equal gaps. The total change from the first term to the fourth term is 20−2=18, so the common difference is d=18/3=6. Therefore x=2+d=8 and y=2+2d=14. Their sum is x+y=8+14=22, so option C is correct. The resulting sequence 2, 8, 14, 20 has consecutive differences 6, 6, and 6, which verifies the calculation. Another method is to write x=2+d and y=2+2d, then use the last term 2+3d=20. The important point is that four terms create three intervals; dividing by four would give the wrong common difference and lead to a distractor.

The original sequence is an arithmetic progression with common difference 11−6=5, also confirmed by 16−11=5. Transform each term t into 2t+3. If two consecutive original terms are t and t+5, their new values are 2t+3 and 2(t+5)+3. Their difference is [2(t+5)+3]−[2t+3]=10. Thus multiplication by 2 scales the original common difference from 5 to 10, while adding the same constant 3 to every term has no effect on differences because the two added constants cancel. The new sequence begins 15, 25, 35, …, whose differences are 10. Therefore option C is correct; 5 ignores the multiplication, 8 does not follow the transformation, and 13 is merely the transformed second term.

The original sequence is an AP with common difference d = 10 − 13 = −3; the next differences confirm this value. Multiplying every term by a constant multiplies the common difference by the same constant. Hence the new difference is d′ = (−2)(−3) = 6. For example, the transformed terms begin −26, −20, −14, −8, and each successive difference is 6. The negative multiplier reverses the order direction and changes the sign of the original difference, so the result is positive. Therefore option B is correct. Option C is the unchanged original difference, while the other options use an incorrect sign or magnitude.

For three consecutive terms in an arithmetic progression, the middle term is the average of the first and third. Therefore b=(a+c)/2=(6+30)/2=36/2=18. The question asks for b−a, not for b itself, so b−a=18−6=12. Equivalently, the total change from a to c is 30−6=24, and this change consists of two equal AP gaps: b−a and c−b. Each gap is therefore 24/2=12. Hence option B is correct. Option C is the middle term b, option D is the complete end-to-end difference c−a, and option A does not follow from the AP condition. The governing idea is that equal consecutive differences divide the total span between three consecutive terms into two equal parts.

The governing concept is that consecutive terms of an arithmetic progression have equal differences. For three consecutive terms, the middle term is the average of the first and third terms. Thus h = (6 + 24) / 2 = 30 / 2 = 15. The required expression is h − 6, so h − 6 = 15 − 6 = 9. The same result can be obtained from the total change: 24 − 6 = 18, and this change is divided into two equal common differences, giving 18 / 2 = 9. Hence the sequence is 6, 15, 24 and option B is correct. Option C is the total change from the first term to the last, not the common difference or the requested expression.

For three quantities to be consecutive terms of an arithmetic progression, the difference between the second and first must equal the difference between the third and second. Equivalently, twice the middle term equals the sum of the first and third terms. Applying this condition gives 2(k + 5) = (k − 2) + (2k + 1). Therefore 2k + 10 = 3k − 1, so k = 11. Substitution verifies the result: the three terms become 9, 16, and 23. Their successive differences are 16 − 9 = 7 and 23 − 16 = 7, so they really form an AP. Hence option D is correct. The other listed values produce unequal successive differences and therefore cannot satisfy the AP condition.

An arithmetic progression is a sequence in which subtraction of each term from the following term gives the same constant value. To test option A, calculate 5 − 9 = −4, 1 − 5 = −4, and −3 − 1 = −4. Thus every consecutive difference is −4, so option A satisfies both requirements. In option B, the differences are −5, −4, and −4, so they are not constant. Option C has a constant difference of +4, making it an AP but not one with difference −4. Option D has differences −4 and −5, so it is not an AP. Therefore option A is the only valid sequence. The negative common difference indicates that the terms decrease by 4 at every step.

An arithmetic progression can be written in the form a_n=a_1+(n-1)d, so the coefficient of n in a linear expression for its terms equals the common difference. Here a_n=7-0.5n. Increasing n by 1 changes the term by -0.5: a_{n+1}-a_n=[7-0.5(n+1)]-[7-0.5n]=-0.5. Thus the sequence decreases by one-half at every step, and its common difference is -0.5. Option D is correct. Option B gives the magnitude but misses the negative sign, while options A and C confuse the constant or the coefficient with the required difference.

For three consecutive terms of an arithmetic progression, the two consecutive differences must be equal. The first difference is (2x + 1) − (x − 2) = x + 3. The second difference is (4x − 3) − (2x + 1) = 2x − 4. Equating them gives x + 3 = 2x − 4, so x = 7. Substitute x = 7 into either difference: d = x + 3 = 7 + 3 = 10. The terms then become 5, 15, and 25, whose differences are both 10, confirming the result. Therefore option C is correct. Every other option fails because its proposed x does not make the two consecutive differences equal.

Use the standard nth-term form of an arithmetic progression, aₙ = a₁ + (n − 1)d. Rewrite the given expression as aₙ = 12 + (n − 1)(−7). Comparing the two forms immediately gives a₁ = 12 and d = −7. Direct substitution verifies both values: when n = 1, a₁ = 12 − 7(0) = 12; when n = 2, a₂ = 12 − 7 = 5, so a₂ − a₁ = 5 − 12 = −7. Therefore option C is correct. Option A loses the negative sign and would describe an increasing sequence. Option B mistakes the second term for the first, while option D interchanges the roles of the first term and common difference.