Arithmetic Progression

For an arithmetic progression, \(a_n=a_1+(n-1)d\). Thus, \(32=12+(6-1)d=12+5d\). Hence \(5d=20\), so \(d=4\). There are 5 gaps, not 6, between \(a_1\) and \(a_6\). Exam tip: while finding the common difference from two terms, use the difference between their term numbers.

Direct answer: option C, 95. The first term is a = 5, the common difference is d = 12−5 = 7, and the number of terms is n = 5. Use the AP sum formula Sₙ = n/2[2a + (n−1)d]. Thus S₅ = 5/2[2(5) + 4(7)] = 5/2[10 + 28] = 5/2 × 38 = 95. We can verify by finding the terms directly: 5, 12, 19, 26, and 33. Their sum is 5 + 12 + 19 + 26 + 33 = 95. Option A is too small, while B and D do not equal the calculated sum. They may result from using the wrong common difference or omitting or miscounting a term. Another check is pairing first and last terms: 5 + 33 = 38 and 12 + 26 = 38, with 19 in the middle; total 38 + 38 + 19 = 95. Count exactly five terms.

The governing concept is the general form of an arithmetic progression. If a_n = 42 − 6n, then increasing n by 1 changes the term by −6: a_(n+1) − a_n = [42 − 6(n+1)] − [42 − 6n] = 42 − 6n − 6 − 42 + 6n = −6. Therefore the common difference is −6, so option A is correct. The negative sign shows that the progression decreases by 6 at each step. Option B ignores that sign, while 36 and 42 are merely numbers appearing through multiplication or the constant term and are not the difference between consecutive terms. Checking values also confirms it: a1 = 36 and a2 = 30, so a2 − a1 = −6.

In an arithmetic progression, the fourth term is reached after three equal gaps from the first term. Thus, \(a_4=a_1+3d\). Here, \(30=18+3d\), so \(3d=12\) and \(d=4\). If the common difference were 3, the fourth term would be \(18+3\times3=27\), not 30. Exam tip: from the first term to the \(n\)th term, there are \(n-1\) common differences.

The first term is \(a=11\) and the common difference is \(d=17-11=6\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{12}=11+(12-1)\times6=11+66=77\). Getting 73 usually results from using an incorrect number of common differences. Exam tip: from the first term to the \(n\)th term, the common difference is added \(n-1\) times.

The \(n\)th term of an arithmetic progression is \(a_n=a+(n-1)d\). Hence, \(a_{10}=7+(10-1)\times4=7+36=43\). Although 43 is correct, there are 9 intervals from the first term to the 10th term, so 4 is added 9 times, not 10 times. Exam tip: use \((n-1)\) when finding the \(n\)th term.

In option C, the first term is 6. The differences between consecutive terms are 3 - 6 = -3, 0 - 3 = -3, and -3 - 0 = -3, so its common difference is -3. Option D has common difference -3, but its first term is 9. Exam tip: To find the common difference of an AP, subtract the first term from the second term.

To find the first term, put n=1. Thus, a₁=7(1)-2=5. Therefore, the correct answer is 5. The number 7 is the coefficient of n, not the first term. Exam tip: To find the first term of a sequence, substitute n=1.

Here, the first term is \(a=16\) and the common difference is \(d=5\). The \(n\)th term is \(a_n=a+(n-1)d\). So, \(16+(n-1)\times5=56\) gives \(n-1=8\), hence \(n=9\). Therefore, 56 is the 9th term. The close distractor, the 8th term, is incorrect because the 8th term is \(16+7\times5=51\). Exam tip: To find the position of a given term in an AP, start with \(a_n=a+(n-1)d\).

The first four terms of the arithmetic progression are 9, 15, 21, and 27. Their sum is \(9+15+21+27=72\). Therefore, 72 is the correct answer. A value such as 78 can result from an incorrect addition. Exam tip: For a small number of terms, direct addition is quickest; for more terms, use \(S_n=\frac{n}{2}[2a+(n-1)d]\).

The nth-term formula for an arithmetic progression is \(a_n=a+(n-1)d\). Therefore, \(a_7=25+(7-1)(-5)=25-30=-5\). Hence, \(-5\) is correct. \(0\) would result from subtracting the common difference only 5 times instead of 6 times. Exam tip: when \(d\) is negative, each successive term decreases.

The direct answer is option C: 5. In an arithmetic progression, the difference between terms \(a_m\) and \(a_k\) equals \((m-k)d\). Here, the sixth term is 34 and the third term is 19, so the increase is \(34-19=15\). From the third term to the sixth term there are three equal gaps: third to fourth, fourth to fifth, and fifth to sixth. Hence \(3d=15\), giving \(d=15/3=5\). Option C is correct. Option A, 3, would give an increase of only 9 across three gaps. Option B, 4, would give 12. Option D, 6, would give 18. The common mistake is dividing by 6 or by the number of terms instead of the number of gaps. The formula \(a_n=a_1+(n-1)d\) leads to the same subtraction result.

Given \(a_n=5n+6\), \(a_3=5(3)+6=21\) and \(a_5=5(5)+6=31\). Therefore, \(a_3+a_5=21+31=52\). An option such as 48 may result from incorrectly handling the \(+6\) term. Exam tip: substitute the value of \(n\) separately to find each required term before adding.

Here, the first term is \(a=70\) and the common difference is \(d=62-70=-8\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(14=70+(n-1)(-8)\) gives \(n-1=7\), so \(n=8\). Therefore, 14 is the 8th term. The 7th term is 22, so it is a close but incorrect option. Exam tip: use a negative common difference for a decreasing AP.

The common difference of an arithmetic progression is found by subtracting a term from the next term. Here, for every pair of consecutive terms, \(3-3=0\); therefore, the common difference is \(0\). The number \(3\) is the value of each term, not the common difference. Exam tip: subtract any term from its next term to check the common difference.

The first term is (19) and (d=4), so (a_n=19+(n-1)4=4n+15). Remember (n-1) while forming the general term.

The direct answer is option D, 52. We know the fourth term is 22 and the common difference is 6. Moving from the fourth term to the ninth term requires \(9-4=5\) equal gaps. Each gap adds 6, so the total increase is \(5\times6=30\). Therefore \(a_9=22+30=52\). Option D is correct. Equivalently, \(a_9=a_4+(9-4)d=22+5(6)=52\). Option A, 46, adds only 24, which corresponds to four gaps and would be \(a_8\). Option B, 48, adds 26, not a whole-number multiple of the common difference from 22. Option C, 50, adds 28, also not the required 30. The important idea is to count the gaps between positions, not the positions themselves. From term 4 to term 9 there are five steps: 4-to-5, 5-to-6, 6-to-7, 7-to-8, and 8-to-9. Memory cue: later term equals known term plus position difference times \(d\).

The first term is (13), and adding (5) each time gives (13,18,23,28). Form the terms directly from the given (a) and (d).

The answer is option D, 12. In an arithmetic progression, the difference between consecutive terms is equal. Therefore, the middle term is the average of the first and third terms: \(2(x+6)=x+2x\). Thus \(2x+12=3x\), so \(x=12\). Checking gives 12, 18, 24, whose common difference is 6. Option A, 6, gives 6, 12, 12, so the differences are not equal. Option B, 8, gives 8, 14, 16, with differences 6 and 2. Option C, 10, gives 10, 16, 20, with differences 6 and 4. Option D works exactly because the middle term is halfway between the other two. Remember: for three consecutive AP terms, twice the middle term equals the first term plus the third term.

The direct answer is option B: the 9th term. The sequence has first term \(a_1=45\) and common difference \(d=-6\), because each term decreases by 6. Its general term is \(a_n=45+(n-1)(-6)=45-6(n-1)\). Compute the nearby terms: \(a_8=45-6(7)=45-42=3\), which is still positive. The next term is \(a_9=45-6(8)=45-48=-3\), which is negative. Therefore the first negative term is the ninth term, option B. Option A, the 8th, is wrong because its value is 3, not negative. Option C, the 10th, is negative but is not the first negative term; the ninth already is negative. Option D, the 11th, is even later and also cannot be the first. Exam cue: calculate the last positive term and immediately inspect the next term; do not choose the first merely small positive term.

For the third term, substitute n=3: \(a_3=11-2(3)=11-6=5\). Therefore, the correct answer is 5. Note that 7 is the second term, since \(a_2=11-2(2)=7\). Exam tip: In an nth-term formula, substitute the position number of the required term for n.

The average of the first three terms is (\frac{6+14+22}{3}=14). In an arithmetic progression the average of three consecutive terms is the middle term.

In an arithmetic progression, the nth-term formula is \(a_n=a_1+(n-1)d\). Thus, \(50=5+(10-1)d=5+9d\). Hence \(9d=45\), so \(d=5\). Option 4 is incorrect because there are 9 common-difference gaps from the first term to the tenth term, not 10. Exam tip: always use \((n-1)d\) in the nth-term formula.

In this arithmetic progression, the first term is \(a=2\) and the common difference is \(d=10-2=8\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{15}=2+(15-1)\times 8=2+112=114\). Hence, \(114\) is correct. \(118\) can result from incorrectly using \(15\times 8\); the formula uses \((n-1)\). Exam tip: identify the first term and common difference before substituting in the nth-term formula.

In (9,18,27,36,\ldots), (9) is added each time. To find the common difference check the difference of two consecutive terms.