Arithmetic Progression

The governing pattern is an arithmetic progression with first term 1 and common difference 3. The rule x=3n+1, with n∈W and 0≤n≤4, produces x=1,4,7,10,13 when n=0,1,2,3,4. Thus it gives exactly the five members of B. Option B produces multiples of 3, option C has difference 2, and option D has difference 4, so option A is correct.

The governing concept is the nth-term formula for an arithmetic progression. The first term is \(a=13\), and the common difference is \(d=5\), since each term increases by 5. Therefore \(a_n=a+(n-1)d=13+5(n-1)\). To find the position of 98, set \(13+5(n-1)=98\). This gives \(5(n-1)=85\), then \(n-1=17\), and finally \(n=18\). Thus 98 is the 18th term, so option C is correct. The expression \(n-1\) is essential because the first term requires zero additions of the common difference. Using \(13+5n\) would shift the position incorrectly.

The governing concept is the arithmetic progression, whose first term is 5 and common difference is 7. Its nth term is a_n = 5 + (n - 1)7 = 7n - 2. We need 50 < 7n - 2 < 150. Adding 2 gives 52 < 7n < 152, and division by 7 gives 7.43... < n < 21.71.... Therefore the integer values of n are 8 through 21. The number of integers in this inclusive range is 21 - 8 + 1 = 14. The corresponding first and last terms are 54 and 145, both satisfying the strict inequalities. Thus option B is correct; options A, C, and D result from miscounting one or more endpoints or treating 50 or 150 as included.

The common difference of an arithmetic progression is the fixed amount by which one term changes to become the next term. It is calculated by subtracting a term from the following term. In this sequence, the first two terms are 5 and 9, so the difference is \(9-5=4\). Checking the next pairs gives \(13-9=4\) and \(17-13=4\), confirming that the difference remains constant.

Therefore the common difference is 4, which is option B. The number 3 in option A is not the change between consecutive terms, and 5 is merely the first term, not the difference. The number 9 is the second term. A reliable method is to subtract the first term from the second and then check one more pair if necessary, especially when identifying an arithmetic progression.

The nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Here, \(a=7\), \(d=3\), and \(n=5\), so \(a_5=7+(5-1)\times3=7+12=19\). Therefore, 19 is correct. Choosing 18 would mean adding 3 only three times, but there are four common differences from the first term to the fifth term. Exam tip: always use \(n-1\) when finding the nth term.

The direct answer is A: yes, because the common difference is -4 . An arithmetic progression is a sequence in which the difference between each pair of consecutive terms is the same. Check the terms step by step: the second term minus the first is 16 - 20 = -4; the third minus the second is 12 - 16 = -4; the fourth minus the third is 8 - 12 = -4. Since the difference remains -4, this is an arithmetic progression, with first term 20 and common difference d=-4 . Option A is correct because it gives the exact reason. Option B is wrong because an arithmetic progression may increase, decrease, or remain constant; decreasing is not a disqualification. Option C is wrong because positivity is not the test: the sequence can later contain zero or negative terms and still be arithmetic. Option D is wrong because the first term can be any number; having first term 20 does not prevent a progression. Exam cue: subtract consecutive terms, and accept the sequence when the results are equal.

The nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Therefore, \(a_4=11+(4-1)(-2)=11-6=5\). Hence, 5 is correct. The answer 7 results from treating the negative common difference incorrectly. Exam tip: always retain the negative sign of \(d\) while substituting in the formula.

The direct answer is option A, the 9th term. This is an arithmetic progression with first term \(a_1=6\) and common difference \(d=10-6=4\). Use \(a_n=a_1+(n-1)d\). Set the required value equal to 38: \(6+4(n-1)=38\). Subtract 6: \(4(n-1)=32\); divide by 4: \(n-1=8\); add 1: \(n=9\). Thus 38 is the ninth term. Option A is correct. Option B, 8th, would give \(6+7(4)=34\). Option C, 10th, would give \(6+9(4)=42\). Option D, 7th, would give \(6+6(4)=30\). The key idea is that the question asks for a position, so solve for \(n\), not for another term. Checking by listing also gives 6, 10, 14, 18, 22, 26, 30, 34, 38. Memory cue: put the known value equal to \(a_n\), then solve for the position.

The governing definition is that an arithmetic progression has the same difference between every pair of consecutive terms. For option B, 5 − 1 = 4, 9 − 5 = 4, and 13 − 9 = 4, so the common difference is constant and the sequence is an arithmetic progression. Therefore option B is correct. Option A doubles each term, showing a constant ratio rather than a constant difference. Option C also doubles each term, so it is geometric, not arithmetic. In option D, the terms are consecutive squares: 1, 4, 9, 16; its differences are 3, 5, and 7, which are not equal. Checking differences, rather than merely observing that the terms increase, is the reliable test.

For an arithmetic progression, \(a_n=a_1+(n-1)d\). Thus, \(a_6=4+5d\). Substituting the given values gives \(24=4+5d\), so \(5d=20\) and \(d=4\). Option 5 is incorrect because there are 5 gaps between the first and sixth terms, not 4. Exam tip: always use \((n-1)d\) in the formula for \(a_n\).

In the linear form (a_n=2n+5), the coefficient of (n) is (2). In an arithmetic progression, this coefficient is the common difference.

Here (a=15) and (d=-3), so (a_7=15+6(-3)=-3). In a decreasing progression, terms may go below zero.

In an arithmetic progression, the difference between consecutive terms must remain constant. Here, 9−5=4, 14−9=5, and 18−14=4, so it is not an AP. Increasing terms alone are not enough. In exams, always check consecutive differences first.

The first term is (8) and the difference is (6), so (a_n=8+(n-1)6=6n+2). In the general term, (d) appears with (n).

In an arithmetic progression, the fourth term is \(a_4=a_1+3d\), because there are three equal gaps between the first and fourth terms. Thus, \(27=12+3d\), so \(3d=15\) and \(d=5\). If the difference were 6, the fourth term would be \(12+3\times6=30\), not 27. Exam tip: use \(a_n=a+(n-1)d\) for the nth term.

The first term is (4), and adding (7) each time gives (4,11,18,25,\ldots). Both the first term and the difference must match.

The direct answer is A: 99. The progression starts with first term a=4 and has common difference d=5 , because 9-4=5, 14-9=5, and 19-14=5. Continuing the pattern gives the first six terms as 4, 9, 14, 19, 24, and 29. Add them carefully: 4+9=13, 13+14=27, 27+19=46, 46+24=70, and 70+29=99. The formula gives the same result: S_n=\frac{n}{2}[2a+(n-1)d]=\frac{6}{2}[2(4)+5(5)]=3(33)=99 . Option A is correct. Option B, 100, is one more than the correct total and may result from an addition error. Option C, 101, is also incorrect because the listed terms total 99. Option D, 105, does not follow from either direct addition or the formula. Always check that six terms, not five or seven, have been used. Memory cue: write the terms first when n is small, then add or verify with the formula.

In this arithmetic progression, the first term is \(a=1\) and the common difference is \(d=6-1=5\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{12}=1+(12-1)\times5=1+55=56\). Hence, 56 is correct. The option 51 results from the common mistake of adding only 10 differences instead of 11. Exam tip: always use \((n-1)\) when finding the \(n\)th term of an AP.

In an arithmetic progression, the second term is \(a_2=a_1+d\). Hence, \(a_1=a_2-d=13-4=9\). Therefore, the correct answer is 9. Choosing 8 would result from an incorrect subtraction. Exam tip: To find the first term from the next term, subtract the common difference.

In (3,6,12,24,\ldots), the differences are (3,6,12), which are not equal. For an arithmetic progression, the difference must always be constant.

The direct answer is option C, the 8th term. The first term is 40 and the common difference is \(35-40=-5\). Therefore \(a_n=40+(n-1)(-5)\). Set the term equal to 5: \(40-5(n-1)=5\). Subtract 40: \(-5(n-1)=-35\); divide by \(-5\): \(n-1=7\); hence \(n=8\). So option C is correct. Option A, 6th, gives \(40-5(5)=15\). Option B, 7th, gives \(40-5(6)=10\). Option D, 9th, gives \(40-5(8)=0\). The negative difference simply means the values fall by 5; it does not make the position negative. Listing confirms: 40, 35, 30, 25, 20, 15, 10, 5. A common error is to count the number of subtractions incorrectly. From the first term to the eighth term there are seven gaps. Memory cue: in a decreasing AP, retain the minus sign and use \(n-1\) gaps.

In an arithmetic progression, the seventh term is \(a_7=a_1+6d\) because there are 6 gaps between the first and seventh terms. Thus, \(33=3+6d\), so \(6d=30\) and \(d=5\). If the common difference were 6, the seventh term would be \(3+6\times6=39\), which is incorrect. Exam tip: In \(a_n=a+(n-1)d\), remember to use \(n-1\), not \(n\).

Given \(a_n=4n-1\). To find the eighth term, substitute \(n=8\): \(a_8=4(8)-1=32-1=31\). Hence, 31 is correct. The value 29 could result from an incorrect multiplication. Exam tip: after substituting the value of \(n\), check multiplication and subtraction separately.

The first four terms of the given arithmetic progression are 7, 10, 13, and 16. Therefore, their sum is \(7+10+13+16=46\). Hence, option C is correct. The value 44 is the sum of only the first three terms, so it is not correct. Exam tip: When the number of terms is small, direct addition is usually the quickest method.

In (18,12,6,0,\ldots), (6) is subtracted each time, so (d=-6). The sign is very important in common difference.