Arithmetic Progression

The nth term of an arithmetic progression is \(a_n=a_1+(n-1)d\). Thus, \(a_{12}=20+(12-1)(-2)=20-22=-2\). Therefore, \(-2\) is correct. The value \(0\) would be obtained for the 10th term, not the 12th term. Exam tip: retain the negative sign of \(d\) carefully in a decreasing AP.

There are (6) differences from the fourth to the tenth term, so (a_{10}=26+6\times6=62). In exams, count the difference in positions.

Here, the first term is \(a=100\) and the common difference is \(d=92-100=-8\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(20=100+(n-1)(-8)\), giving \(8(n-1)=80\) and hence \(n=11\). Therefore, 20 is the 11th term. The 10th term is 28, so it is a close but incorrect option. Exam tip: keep the common difference negative for a decreasing arithmetic progression.

From (114=\frac{6}{2}[8+5d]), (d=6). In exams, keep (2a) and ((n-1)d) correct in the (S_n) formula.

Here, the first term is \(a=14\) and the common difference is \(d=5\). Using \(a_n=a+(n-1)d\), we get \(a_9=14+8\times5=54\) and \(a_{11}=14+10\times5=64\). Therefore, \(a_9+a_{11}=54+64=118\). Option 124 may result from incorrectly counting the number of differences for \(a_{11}\). Exam tip: always use \((n-1)d\) for the \(n\)th term.

Given a_n=50-3n and a_n=5, we get 50-3n=5. Thus, 3n=45 and n=15. Therefore, the 15th term is equal to 5. For the 14th term, the value is 50-3(14)=8, so it is not correct. Exam tip: To find a term number, equate the given term value to a_n and solve for n.

This is the sum of the first (9) multiples of (7), which is (315). In exams, identify sequences of multiples quickly.

In an AP, the difference between two terms equals the difference in their positions multiplied by the common difference: \(a_{12}-a_5=(12-5)d\). Thus, \(72-30=7d\), so \(42=7d\) and \(d=6\). Choosing 7 would confuse the difference in term positions with the common difference. Exam tip: when using \(a_n=a+(n-1)d\), track the term numbers carefully.

For \(a_n=7-3n\), \(a_{n+1}-a_n=[7-3(n+1)]-(7-3n)=-3\), which is constant for every \(n\). Hence it is an AP. In \(n^2-3\), the difference changes. Exam tip: always test consecutive-term differences.

Using \(T_n=a+(n-1)d\), we get \(T_p+T_r=2a+(p+r-2)d\). Since \(p+r=2q\), this becomes \(2[a+(q-1)d]=2T_q\). The product relation is not generally true. Exam tip: terms equally spaced in an AP have the middle term as their average.

Given \(a_n=6n+1\), \(a_4=6(4)+1=25\) and \(a_8=6(8)+1=49\). Therefore, \(a_4+a_8=25+49=74\). Option 72 may result from ignoring the constant term \(+1\), which is incorrect. Exam tip: substitute the correct value of \(n\) each time you find a term.

Here, the first term is 22 and the common difference is 5. Thus, the nth term is \(a_n=22+(n-1)\times5\). Substituting \(a_n=87\), we get \(22+5(n-1)=87\), so \(5(n-1)=65\), \(n-1=13\), and \(n=14\). Therefore, 87 is the 14th term of the AP. If 13 is chosen, the term would be \(82\), not 87. Exam tip: while finding a term number, handle the \(n-1\) carefully.

Write the terms as \(p, p+d, p+2d, p+3d\). Then \(p+s=p+(p+3d)=2p+3d\), while \(q+r=(p+d)+(p+2d)=2p+3d\). Hence A is correct; \(ps=qr\) is not generally true. Exam tip: the sum of outer terms equals the sum of inner terms.

Here, the first term is \(a=8\), the common difference is \(d=12-8=4\), and \(n=25\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), \(S_{25}=\frac{25}{2}[2(8)+24(4)]=\frac{25}{2}(112)=1400\). Therefore, 1400 is correct. An option such as 1350 may result from an error in finding the last term or in using \((n-1)\). Exam tip: for \(n\) terms, always use \((n-1)\) with the common difference in the sum formula.

The formula for the nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Therefore, \(a_6=a+5d\). Substituting the given values gives \(41=11+5d\), so \(5d=30\) and \(d=6\). If 5 were chosen, the sixth term would be \(11+5\times5=36\), not 41. Exam tip: the coefficient of \(d\) in \(a_n\) is always \(n-1\), not \(n\).

Here, the first term is \(a=13\) and the common difference is \(d=10-13=-3\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(13+(n-1)(-3)=-8\), giving \(-3(n-1)=-21\). Hence \(n-1=7\) and \(n=8\). Therefore, \(-8\) is the eighth term. The ninth term would be \(-11\), so it is not correct. Exam tip: write the negative common difference carefully before applying the \(n\)th-term formula.

(a_5=38) and (a_6=46), so the sum is (84). In exams, find both terms separately and add them.

In (S_n=\frac{n}{2}[12+5(n-1)]), putting (n=11) gives (451). In exams, you can also check quickly by substituting options.

For three consecutive terms of an arithmetic progression, the middle term is the arithmetic mean of the first and third. Therefore 2(5x + 2) = (4x - 1) + (7x - 3). Expanding gives 10x + 4 = 11x - 4, so x = 8. Substitution verifies the terms: 4(8) - 1 = 31, 5(8) + 2 = 42, and 7(8) - 3 = 53; the two successive differences are 11 and 11. Hence option D is correct. The original option set did not contain 8 and its marked answer was inconsistent, so the final option set has been corrected. Values 4, 5, and 6 fail the equal-difference condition.

Consecutive AP terms have common difference \(d\). Thus \(q=p+d\) and \(r=p+2d\), so \(p+r=2p+2d=2q\). Option A is not generally true. Exam tip: the middle term is the average of the end terms.

Direct answer: Option B, 24. The first term is \\(a=96\\), and the common difference is \\(d=88-96=-8\\). The tenth term formula is \\(a_n=a+(n-1)d\\). For the tenth term, there are nine differences after the first term, so \\(a_{10}=96+(10-1)(-8)=96-72=24\\). Listing the sequence also confirms it: 96, 88, 80, 72, 64, 56, 48, 40, 32, 24. Option A, 16, would require one extra subtraction of 8 and is actually the eleventh term. Option B, 24, is correct. Option C, 32, is the ninth term, not the tenth. Option D, 40, is the eighth term. The negative sign matters because the progression decreases. Common mistake: using ten subtractions instead of nine; the first term already occupies position one, so term 10 is nine steps away.

To find which term equals 121, equate the general term to 121: \(7n-5=121\). Thus, \(7n=126\), so \(n=18\). Therefore, 121 is the 18th term of the sequence. The 17th term is \(114\), so it is not correct. Exam tip: To find the position of a given value, set \(a_n\) equal to that value.

For an arithmetic progression, the nth-term formula is \(a_n=a+(n-1)d\). Here, \(93=18+(16-1)d\), so \(93=18+15d\). Thus, \(15d=75\) and \(d=5\). The value 6 can result from incorrectly counting the number of gaps between the first and sixteenth terms. Exam tip: there are always \(n-1\) gaps from the first term to the nth term.

The middle term is (x=\frac{10+34}{2}=22). In exams, the middle term of three AP terms is the average.

In an arithmetic progression, the fourth term is \(a_4=a+(4-1)d=a+3d\). Thus, \(29=a+3(6)=a+18\), so \(a=11\). If 13 were chosen, the fourth term would be \(13+18=31\), not the given 29. Exam tip: use \(a_n=a+(n-1)d\) for the \(n\)th term.