Arithmetic Progression

To find the position of 107, put a_n = 107 in the general term: 6n + 5 = 107. Thus, 6n = 102 and n = 17. Therefore, 107 is the 17th term. The 16th term is 6(16) + 5 = 101, so it is not correct. Exam tip: To find the term number for a given value, equate a_n to that value.

The nth-term formula of an arithmetic progression is \(a_n=a+(n-1)d\). Here, \(a=21\) and \(a_{18}=106\), so \(106=21+17d\). Thus, \(17d=85\), giving \(d=5\). If the common difference were 4, the eighteenth term would be \(21+17\times4=89\), not 106. Exam tip: from the first term to the nth term, the common difference is added \(n-1\) times, not n times.

The governing concept is the sum of the first n terms of an arithmetic progression: S_n = n/2 [2a + (n - 1)d]. In this sequence, a = 6, d = 11 - 6 = 5, and n = 13. Thus S_13 = 13/2 [2(6) + (13 - 1)(5)] = 13/2 [12 + 60] = 13/2 × 72 = 13 × 36 = 468. Option A is therefore correct. The same result can be checked by finding the last term, 6 + 12×5 = 66, and using average × number of terms: (6 + 66)/2 × 13 = 36×13 = 468. The other options arise from arithmetic or formula errors.

The middle term is (x=\frac{14+42}{2}=28). In exams, the middle term of three AP terms is the average.

Use the arithmetic-progression sum formula S_n = n/2 [2a + (n - 1)d]. The first term is a = 10, the common difference is d = 17 - 10 = 7, and the required number of terms is n = 8. Therefore S_8 = 8/2 [2(10) + (8 - 1)(7)] = 4[20 + 49] = 4×69 = 276. Hence option B is correct. A useful check is to find the eighth term: a_8 = 10 + 7×7 = 59. The average of the first and last terms is (10 + 59)/2 = 34.5, and 8×34.5 = 276. The other choices do not satisfy this calculation.

In an AP, consecutive differences are equal: \(q-p=r-q\). Rearranging gives \(2q=p+r\), so the middle term is the average of the other two. \(q^2=pr\) is the condition for a GP. Exam tip: check the middle term.

The rule (a_n=50-6n) gives (44) at (n=1) and (38) at (n=2). In exams, match the first term in a decreasing progression.

Use the AP formula \(a_n=a+(n-1)d\). Thus, \(a_8=15+(8-1)d=15+7d\). Since \(a_8=71\), \(71=15+7d\), so \(7d=56\) and \(d=8\). Do not choose \(7\): there are seven common differences from the first term to the eighth term. Exam tip: write \((n-1)\) explicitly before substituting values in the AP formula.

Here, the first term is \(a=5\) and the common difference is \(d=9\). The \(n\)th term is \(a_n=a+(n-1)d\). So, \(5+(n-1)\times9=95\) gives \((n-1)\times9=90\), hence \(n=11\). Therefore, 95 is the eleventh term. The tenth term is 86, so it is not correct. Exam tip: when asked for a term number, equate \(a_n\) to the given term and solve for \(n\).

For an arithmetic progression, the nth term is \(a_n=a+(n-1)d\). Thus, \(a_9=19+(9-1)(-5)=19-40=-21\). Therefore, option B is correct. The value \(-19\) would result from incorrectly using 7 in place of \(n-1\). Exam tip: when the common difference is negative, check the sign after multiplication carefully.

In an AP, consecutive differences are equal: \(q-p=r-q\). Rearranging gives \(2q=p+r\), so the middle term is the average of the outer terms. Exam tip: double the middle term to test three terms quickly.

Here, the first term is \(a=3\), the common difference is \(d=7-3=4\), and \(n=20\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), \(S_{20}=\frac{20}{2}[2(3)+19(4)]=10(82)=820\). Therefore, 820 is correct. A value such as 810 can result from using an incorrect last term or replacing \((n-1)d\) incorrectly. Exam tip: for the sum of \(n\) terms, use \((n-1)d\) in the formula.

(a_{10}-a_4=6d=42), so (d=7). In exams, divide the difference of terms by the difference of positions.

In an arithmetic progression, \(a_n=a_1+(n-1)d\). Hence, \(a_7=a_1+6d\). Substituting the given values, \(52=a_1+6\times7=a_1+42\), so \(a_1=10\). If 12 were used, the seventh term would be \(12+42=54\), not 52. Exam tip: the number of common differences before the \(n\)th term is always \(n-1\).

The position difference is (10) and (d=8), so the difference is (10\times8=80). In exams, use ((m-n)d) directly for the difference of two terms.

The first eight terms are (7,17,27,37,47,57,67,77), and their sum is (336). In exams, you can verify the sum by listing terms from the rule.

Here, the first term is \(a=23\) and the common difference is \(d=31-23=8\). The \(n\)th term is \(a+(n-1)d\). Thus, \(23+(n-1)\times 8=111\) gives \((n-1)=11\), so \(n=12\). Therefore, 111 is the 12th term. The 11th term is \(23+10\times8=103\), so it is not correct. Exam tip: identify \(a\) and \(d\) first, then equate the given value to \(a_n\).

For an arithmetic progression, \(a_n=a+(n-1)d\). Thus, \(a_2=a+d=16\) and \(a_9=a+8d=65\). Subtracting the two equations gives \(7d=49\), so \(d=7\). Substituting this into \(a+d=16\) gives \(a=9\). Therefore, the correct answer is 9. Option 8 can result from an incorrect substitution after finding the common difference. Exam tip: form equations for the given terms and subtract them first to find \(d\) quickly.

Here, the first term is \(a=64\) and the common difference is \(d=56-64=-8\). The \(n\)th term is \(a_n=a+(n-1)d\). For the zero term, \(64+(n-1)(-8)=0\), so \(n-1=8\) and hence \(n=9\). Therefore, the ninth term is zero. The eighth term is \(8\), not zero. Exam tip: To find the position of a specified term, write the \(a_n\) formula first and substitute the given value.

The formula for an arithmetic progression is \(a_n=a+(n-1)d\). Hence, \(a_{12}=a+11d\). Substituting the given values, \(88=11+11d\), so \(11d=77\) and \(d=7\). Therefore, option C is correct. If \(d=8\), the twelfth term would be \(99\), not the given value. Exam tip: the coefficient of \(d\) in the \(n\)th term is always \(n-1\).