Arithmetic Progression

Correct answer: B. 10

Explanation: 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 \(n-1\).

Correct answer: B. (80)

Explanation: 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.

Correct answer: C. (336)

Explanation: 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.

Correct answer: C. 12th term

Explanation: 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 \(a_n\).

Correct answer: C. 9

Explanation: 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.

Correct answer: C. 9th term

Explanation: 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.

Correct answer: C. 7

Explanation: 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 \(n-1\).

Correct answer: C. 1050

Explanation: Here, the first term is \(a=14\), the common difference is \(d=22-14=8\), and \(n=15\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), \(S_{15}=\frac{15}{2}[2(14)+14(8)]=\frac{15}{2}(140)=1050\). Hence, 1050 is correct. A value such as 1040 usually results from an addition or multiplication error. Exam tip: calculate \(n-1\) first before substituting in the formula.

Correct answer: C. 14

Explanation: Here, the first term is \(a=7\) and the common difference is \(d=5\). The \(n\)th term is \(a_n=a+(n-1)d\). So, \(72=7+(n-1)\times5\), giving \(65=5(n-1)\), hence \(n-1=13\) and \(n=14\). Therefore, 72 is the 14th term of the AP. The 13th term is 67, so 13 is not correct. Exam tip: When the last term is given, equate it to \(a_n\) and solve for \(n\).

Correct answer: B. 84

Explanation: Here, \(S_4\) denotes the sum of the first four terms: \(9+17+25+33=84\). Therefore, the correct answer is 84. The value 80 is obtained by adding only the first and last terms, so it is not the required sum. Exam tip: when only a few terms are given, direct addition is quick and reliable.

Correct answer: C. 16th term

Explanation: Here, the first term is \(a=18\) and the common difference is \(d=25-18=7\). The \(n\)th term is \(a_n=a+(n-1)d\). So, \(18+(n-1)\times7=123\), giving \((n-1)\times7=105\). Hence \(n-1=15\) and \(n=16\). Therefore, 123 is the 16th term. The 15th term is \(18+14\times7=116\), so it is not correct. Exam tip: after finding \(n-1\), remember to add 1 to get the term number.

Correct answer: A. \(2q=p+r\)

Explanation: 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. \(q^2=pr\) is the corresponding condition for a GP. Exam tip: check the average-middle-term property quickly.

Correct answer: C. (722)

Explanation: (S_{14}=\frac{14}{2}[12+13\times7]=721). In exams, calculate the bracket and multiplication carefully.

Correct answer: B. (101)

Explanation: There are (6) differences from the sixth to the twelfth term, so (a_{12}=47+6\times9=101). In exams, count the difference in positions.

Correct answer: A. 9th term

Explanation: Here, the first term is \(a=84\) and the common difference is \(d=75-84=-9\). Using \(a_n=a+(n-1)d\), we get \(12=84+(n-1)(-9)\). Thus, \(n-1=8\), so \(n=9\). Therefore, 12 is the ninth term. The tenth term is 3, so the closest distractor is not correct. Exam tip: always use a negative common difference for a decreasing AP.

Correct answer: B. (6)

Explanation: From (216=\frac{8}{2}[12+7d]), (d=6). In exams, keep (2a) and ((n-1)d) correct in the (S_n) formula.

Correct answer: C. 182

Explanation: Here, the first term is \(a=21\) and the common difference is \(d=28-21=7\). Using \(a_n=a+(n-1)d\), we get \(a_9=21+8\times7=77\) and \(a_{13}=21+12\times7=105\). Therefore, \(a_9+a_{13}=77+105=182\). Option 168 may result from using an incorrect difference instead of \((n-1)d\). Exam tip: while finding the \(n\)th term of an AP, always use \((n-1)d\).

Correct answer: B. 11th term

Explanation: Given a_n=74-5n and a_n=19. So, 74-5n=19, which gives 5n=55 and n=11. Hence, 19 is the 11th term. For example, the 12th term is 74-5(12)=14, not 19. Exam tip: To find a term number, equate the given term value to a_n and solve for n.

Correct answer: C. (624)

Explanation: This is the sum of the first (12) multiples of (8), which is (624). In exams, identify sequences of multiples quickly.

Correct answer: C. \(7\)

Explanation: For an arithmetic progression, \(a_n=a+(n-1)d\). Therefore, \(a_{14}-a_6=(14-6)d=8d\). Using the given values, \(94-38=56=8d\), so \(d=7\). If \(d=6\), the difference across 8 positions would be only \(48\), not the required \(56\). Exam tip: when two terms are given, divide the difference of their values by the difference of their term numbers.

Correct answer: B. 4

Explanation: In an AP, the difference between two terms equals the common difference times the gap in their term numbers. Here the terms are 6 positions apart and 49−25=24, so d=24/6=4. Exam tip: always subtract the term indices as well as the term values.

Correct answer: A. \(2q=p+r\)

Explanation: Consecutive AP terms have equal differences, so \(q-p=r-q\). Rearranging gives \(2q=p+r\), meaning the middle term is the average of the outer terms. Option B does not express this. Exam tip: test the middle term first.

Correct answer: B. 13

Explanation: Here, the first term is \(a=27\) and the common difference is \(d=35-27=8\). The nth term is \(a_n=a+(n-1)d\). Thus, \(123=27+(n-1)8\), so \((n-1)8=96\), \(n-1=12\), and \(n=13\). Option 12 is incorrect because it is the value of \(n-1\), not the term number \(n\). Exam tip: when finding the term number, remember to add 1 after calculating \(n-1\).

Correct answer: A. \(x,\ 2x,\ 3x,\ 4x,\ldots\)

Explanation: In option A, consecutive differences are \(2x-x=x\) and \(3x-2x=x\), so the common difference is fixed. Option D has changing differences. Exam tip: compare two successive differences to identify an AP.

Correct answer: C. 2100

Explanation: Here, the first term is \(a=12\), the common difference is \(d=18-12=6\), and the number of terms is \(n=25\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), \(S_{25}=\frac{25}{2}[2(12)+24\times6]=\frac{25}{2}(168)=2100\). Therefore, option C is correct. Taking \(24\) as the first term would be incorrect; it is the value of \((n-1)\). Exam tip: always check \(n-1\) carefully in the sum formula.