Arithmetic Progression

The direct answer is option D, 32. Compare consecutive terms: \(17-12=5\), \(22-17=5\), and \(27-22=5\). Thus the common difference is 5. Add it to the last shown term: \(27+5=32\). Option D is correct. Option A, 29, would increase by only 2; option B, 30, would increase by 3; and option C, 31, would increase by 4. None preserves the common difference of 5. This is an arithmetic progression, so the general rule would be \(a_n=12+(n-1)5\), which also predicts the next term when \(n=5\): \(12+4(5)=32\). The safest method is to calculate the differences rather than assume the numbers are being rounded or changed irregularly. Memory cue: in a constant-step sequence, the next term is the last term plus the repeated step.

In an arithmetic progression, each next term is obtained by adding the common difference to the previous term. Therefore, the second term is 6 + 7 = 13. The value 14 would result from adding 7 twice. Exam tip: remember that the second term is always a + d.

Each term decreases by (5), so the common difference is (-5). In exams, write the difference as negative for a decreasing progression.

The defining concept of an arithmetic progression is that the difference between every pair of consecutive terms must be constant. Calculate the differences: 12 − 6 = 6, 18 − 12 = 6, and 24 − 18 = 6. Since all these consecutive differences are equal, the sequence is an arithmetic progression with common difference 6. Therefore option A is correct. Option B confuses the second term, 12, with the common difference. Option C is false because the calculated differences are equal. Option D is also false because the terms increase rather than decrease. An arithmetic progression can have a positive, negative, or zero common difference; this particular sequence has the positive difference 6.

For the given sequence, the consecutive differences are \(5-2=3\), \(10-5=5\), and \(17-10=7\). Since these differences are not equal, it is not an arithmetic progression. Merely having increasing terms does not make a sequence an arithmetic progression. Exam tip: always check the difference between consecutive terms to identify an AP.

The governing concept is the meaning of the first term in an arithmetic progression. The first term is the number written first in the ordered list, and it is commonly represented by a. In the progression 18, 23, 28, 33, ... the first entry is 18, so a = 18 and option B is correct. The number 5 can be found from 23 − 18 = 5, but it is the common difference, not the first term. The numbers 23 and 28 are the second and third terms respectively. Therefore, options A, C, and D confuse either the common difference or a later term with the requested first term. The position in which a value appears is the key to answering this direct question.

The nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Therefore, \(a_3=8+(3-1)\times3=8+6=14\). Option 11 is only the second term because the common difference is added once. Exam tip: for the third term, add \(2d\) to the first term.

The direct answer is option C: 19. This is an arithmetic progression because the difference between consecutive terms is constant: \(28-31=-3\), \(25-28=-3\), and \(22-25=-3\). Therefore the common difference is \(d=-3\). To get the next term, subtract 3 from the last displayed term: \(22-3=19\). Option C is correct. Option A, 17, subtracts 5 and does not follow the established difference. Option B, 18, subtracts 4, so it is also inconsistent. Option D, 20, subtracts only 2. A decreasing arithmetic progression still uses a signed difference; the negative sign must not be ignored. The general formula would be \(a_n=31+(n-1)(-3)\), which also predicts the fifth term \(31+4(-3)=19\). Exam cue: compare consecutive terms and carry the negative sign carefully.

The difference between (17) and (23) is (6), so the missing term is (17-6=11). In exams, apply the difference backward too.

This is a sequence of multiples of (6), so (a_n=6n). In exams, identify multiple-based progressions as (kn).

Given \(a_n=4n+2\). To find the fifth term, substitute \(n=5\): \(a_5=4\times5+2=20+2=22\). Hence, the correct answer is \(22\). The value \(20\) results if the \(+2\) is omitted. Exam tip: substitute the term number carefully for \(n\) in the nth-term formula.

For this arithmetic progression, the first term is \(a=7\) and the common difference is \(d=11-7=4\). Thus, \(a_n=a+(n-1)d=7+4(n-1)=4n+3\). Hence, \(a_n=4n+3\) is correct. In \(a_n=4n-3\), the first term is \(1\), so it cannot represent the given progression. Exam tip: substitute \(n=1\) to check whether a general term gives the first term.

In an arithmetic progression, the second term equals the first term plus the common difference. Thus, \(16+(-4)=12\). Getting 14 would result from an incorrect operation with the common difference. Exam tip: when the common difference is negative, each next term is smaller than the preceding term.

In an arithmetic progression, the difference between every pair of consecutive terms must be the same. Here the differences are \(6-3=3\), \(12-6=6\), and \(24-12=12\). Since these differences are unequal, the sequence is not an arithmetic progression. Merely increasing terms or terms that are multiples of 3 do not make a sequence an AP. Exam tip: Find two or three consecutive differences to check whether a sequence is an AP.

In an arithmetic progression, the common difference \(d\) is the difference between consecutive terms. Here, \(d=10-4=6\); it is confirmed by \(16-10=6\). \(10\) is the second term, not the common difference. Exam tip: use \(d=a_2-a_1\) and verify it with the next pair of terms.

The \(n\)th term of an arithmetic progression is \(a_n=a+(n-1)d\). For the fourth term, \(a_4=15+(4-1)\times4=15+12=27\). Hence, \(27\) is correct. \(31\) would be the fifth term because it requires adding the common difference four times. Exam tip: for the fourth term, add \(3d\) to the first term.

The governing concept is the nth-term formula for an arithmetic progression, in which the difference between consecutive terms is constant. Here, 45−50 = −5, 40−45 = −5, and 35−40 = −5, so the first term is a₁ = 50 and the common difference is d = −5. Use aₙ = a₁ + (n−1)d. For n = 6, a₆ = 50 + (6−1)(−5) = 50 − 25 = 25. Hence option B is correct. Listing the sequence confirms the count: 50 is first, 45 second, 40 third, 35 fourth, 30 fifth, and 25 sixth. Option C is the fifth term, option D is the fourth term, and option A is the seventh term. These distractors arise from shifting the position or stopping the count too early.

The common difference is (7), so the next term is (20+7=27). In exams, continue the same difference for the blank term.

The governing concept is the common difference of an arithmetic progression. It is calculated by subtracting a term from the next term: d = second term − first term. Here d = 15 − 19 = −4. Checking the next pairs gives 11 − 15 = −4 and 7 − 11 = −4, confirming that the difference is constant. Therefore, option B is correct. The negative sign is essential because the progression decreases by 4 at every step; saying +4 describes only the size of the decrease, not the signed common difference. Option C is simply a term of the progression, while option D has the wrong magnitude. A consistent negative difference identifies a decreasing arithmetic progression.

The governing concept is the constant-difference test for an arithmetic progression. Subtract consecutive terms: 28 − 14 = 14, 42 − 28 = 14, and 56 − 42 = 14. Since every difference is 14, the sequence is an arithmetic progression with common difference d = 14. Hence option A is correct. Option B mistakes the second term, 28, for the common difference. Option C is incorrect because an arithmetic progression need not have equal terms; it needs equal differences between consecutive terms. Option D is incorrect because the terms are increasing, not decreasing. The sequence can also be written using the nth-term rule aₙ = 14 + (n − 1)14, which reinforces that the same amount is added at each step. Equal consecutive differences are the decisive reason.

In an arithmetic progression, each new term is obtained by adding the common difference d to the preceding term. Here, the first term is 4 and d = 9: 4, 4+9 = 13, 13+9 = 22, 22+9 = 31. Therefore, (4, 13, 22, 31) is correct. In option C, 9 is incorrectly taken as the first term instead of 4. Exam tip: check that the difference between every pair of consecutive terms equals d.

At (n=1) it gives (36), and at (n=2) it gives (30), so (a_n=42-6n). In exams, check a decreasing rule with the first term.

Given \(a_n=42-6n\). Substituting \(n=5\), we get \(a_5=42-6(5)=42-30=12\). Therefore, 12 is the correct option. Getting 14 would result from an error in subtraction. Exam tip: substitute the value of \(n\) first, then perform multiplication and subtraction in order.

In an arithmetic progression, \(a\) is the first term. Here, the first term is \(5\), so \(a=5\). The common difference \(d\) is the difference between consecutive terms: \(14-5=9\) (also \(23-14=9\)). Therefore, \(d=9\). The option \(a=9,\ d=5\) incorrectly interchanges the first term and the common difference. Exam tip: find \(d\) by subtracting the first term from the second term.

In an AP, \(d=a_{n+1}-a_n\) remains the same for every pair of consecutive terms, so A is correct. A constant ratio is a GP property. Exam tip: compare two consecutive differences.