Arithmetic Progression

The first term is \(a=31\), and the common difference is \(d=35-31=4\). Hence, \(a_n=a+(n-1)d=31+4(n-1)=4n+27\). Therefore, option A is correct. In option C, putting \(n=1\) gives 35, whereas the first term must be 31. Exam tip: Always substitute \(n=1\) in a proposed general term to verify the first term.

The increase over three gaps is (12), so (d=4), then (a_1=19-3(4)=7). From middle terms first find the common difference.

In (6,12,18,24,30,\ldots), the fifth term is (30) and the common difference is (6). Check each option up to the fifth term.

Direct answer: option B, 5. The nth term of an AP is aₙ = a₁ + (n − 1)d, where d is the constant common difference. From a₃ = 15, we get a₁ + 2d = 15. From a₇ = 35, we get a₁ + 6d = 35. Subtracting the first equation from the second gives 4d = 20, so d = 5. Substituting this into a₁ + 2d = 15 gives a₁ + 10 = 15, hence a₁ = 5. The sequence is therefore 5, 10, 15, 20, 25, 30, 35. A (3), C (7), and D (9) do not produce both the third and seventh terms with one fixed difference. The important point is that the gap between the seventh and third terms contains four equal steps, not six.

The governing concept is the general term of an arithmetic progression, aₙ = a₁ + (n − 1)d. From the third term, a₁ + 2d = 11, and from the ninth term, a₁ + 8d = 47. Subtracting the first equation from the second eliminates a₁ and gives 6d = 36, so d = 6. Substituting this value into a₁ + 2d = 11 gives a₁ + 12 = 11, hence a₁ = −1. A direct check gives the sequence −1, 5, 11, 17, …, and its ninth term is −1 + 8(6) = 47. Therefore option A is correct. Values 0, 1, and 2 do not satisfy both given term conditions with one common difference.

The governing concept is the general term of an arithmetic progression, aₙ = a₁ + (n − 1)d. Substituting a₁ = 12 and d = −3 gives aₙ = 12 − 3(n − 1). The first eight terms are 12, 9, 6, 3, 0, −3, −6, and −9. A positive number must be strictly greater than zero, so the positive terms are 12, 9, 6, and 3. There are exactly four of them, making option B correct. Zero is neither positive nor negative and must not be counted. The later terms are negative, so options C and D overcount by including zero or a negative term. Option A undercounts by omitting one of the four positive terms. Listing the terms also confirms the result without relying on an ambiguous sign convention.

For an arithmetic progression, moving from the third term to the seventh term involves four equal common-difference steps. Therefore a₇ − a₃ = 4d. Using the given values, 32 − 12 = 20 = 4d, so d = 5. Since the third term is 12, move backward two steps to obtain the first term: a₁ = 12 − 2(5) = 2. The sequence is consequently 2, 7, 12, 17, 22, 27, 32, which is option B. Option A has common difference 5 but its third term is 10. Option C has third term 12 but common difference 4, and option D has third term 18. Both conditions must hold simultaneously, so checking only one term is insufficient. Hence B is uniquely correct.

The governing concept is the sum formula for the first n terms of an arithmetic progression: S_n = n/2[2a_1 + (n - 1)d]. Here the first term is a_1 = 2, the common difference is d = 8 - 2 = 6, and n = 7. Therefore S_7 = 7/2[2(2) + (7 - 1)6] = 7/2(4 + 36) = 7/2 x 40 = 140. The terms themselves are 2, 8, 14, 20, 26, 32, and 38, whose total is also 140. Hence option C is correct. Pairing the first and last terms gives 2 + 38 = 40, and the three symmetric pairs also total 40, with the middle term 20, giving 3(40) + 20 = 140. The other choices reflect calculation or term-counting errors.

For an arithmetic progression, the nth-term formula is aₙ = a₁ + (n − 1)d. The ninth term and fourth term are five positions apart, so a₉ − a₄ = 5d. Substituting the given values gives 57 − 22 = 35 = 5d, hence d = 7. Now use the fourth-term equation: a₄ = a₁ + 3d. Thus 22 = a₁ + 3(7) = a₁ + 21, so a₁ = 1. Therefore, option A is correct. Checking gives the progression 1, 8, 15, 22, …, and after five more steps the ninth term is 57. Options B, C, and D do not satisfy both given terms simultaneously; they arise from mishandling the number of intervals or subtracting the wrong quantity.

Direct answer: option A, −1. Use aₙ = a₁ + (n − 1)d. The difference between the tenth and third terms covers 10 − 3 = 7 equal steps, so 62 − 13 = 7d. Hence 49 = 7d and d = 7. Now use the third-term equation: 13 = a₁ + 2d = a₁ + 14. Therefore a₁ = 13 − 14 = −1. The resulting progression begins −1, 6, 13, 20, and its tenth term is −1 + 9 × 7 = 62. B (0), C (1), and D (2) would give different third terms when the common difference is 7, so they cannot satisfy both conditions. The key counting idea is that the index gap 10 − 3 determines the number of steps.

An explicit rule gives a term when a value of n is substituted. For n = 1, 2, 3, 4, and 5, the rule aₙ = 4n + 7 produces 11, 15, 19, 23, and 27. Adding these terms gives 11 + 15 + 19 + 23 + 27 = 95, so option C is correct. The same result follows from the AP sum formula. Here the first term is 11, the common difference is 4, and S₅ = 5/2[2(11) + (5 − 1)4] = 5/2(22 + 16) = 5/2 × 38 = 95. The other options could result from omitting a term, using an incorrect value for the first term, or making an addition error. Both methods independently confirm the answer.

Direct answer: option B, 5. For an arithmetic progression, the nth term is aₙ = a₁ + (n − 1)d, where a₁ is the first term and d is the common difference. For the fifth term, a₅ = a₁ + 4d = 33. For the eleventh term, a₁₁ = a₁ + 10d = 75. Subtract the first equation from the second: (a₁ + 10d) − (a₁ + 4d) = 75 − 33, so 6d = 42 and d = 7. Substitute d = 7 into a₁ + 4d = 33: a₁ + 28 = 33, hence a₁ = 5. Thus B is correct. Verification gives a₅ = 5 + 4 × 7 = 33 and a₁₁ = 5 + 10 × 7 = 75. A, 3, would give a fifth term of 31; C, 7, would give 35; D, 9, would give 37, when the common difference is 7, so none satisfies both given terms. The important idea is that the index difference 11 − 5 = 6 represents six equal steps, not five or eleven. A helpful memory cue is: the coefficient of d is always one less than the term number.

Direct answer: option A, −1. Apply aₙ = a₁ + (n − 1)d. The fourth-term condition is a₁ + 3d = 20, and the eleventh-term condition is a₁ + 10d = 69. Subtracting the first equation from the second gives 7d = 49, so d = 7. Substituting back, a₁ + 3(7) = 20, so a₁ + 21 = 20 and a₁ = −1. Check: the fourth term is −1 + 21 = 20, while the eleventh is −1 + 70 = 69. B (0), C (1), and D (2) do not satisfy both stated terms with d = 7. Notice that the index gap 11 − 4 is seven, which explains the coefficient of d when the equations are subtracted.

The governing concept is the nth-term formula for an arithmetic progression: a_n = a_1 + (n - 1)d. For the fifth term, there are four common differences between a_1 and a_5, so a_5 = a_1 + 4d. Substituting the given values gives 28 = a_1 + 4(7) = a_1 + 28. Hence a_1 = 28 - 28 = 0, so option A is correct. The same result can be checked by moving backward from the fifth term and subtracting 7 four times: 28, 21, 14, 7, 0. A frequent mistake is to use five differences instead of four; the number of differences is n - 1 because the first term is the starting point. Options B, C, and D therefore do not produce a fifth term of 28 when d = 7.