Arithmetic Progression

Correct answer: B. (a_n=36-5n)

Explanation: The rule (a_n=36-5n) gives (31) at (n=1) and (26) at (n=2). In exams, match the first term in a decreasing progression.

Correct answer: C. 7

Explanation: The formula for an arithmetic progression is \(a_n=a+(n-1)d\). Hence, \(a_7=12+(7-1)d=12+6d\). So, \(54=12+6d\), giving \(6d=42\) and \(d=7\). If \(d=6\), the seventh term would be \(12+6\times6=48\), not 54. Exam tip: use \(n-1\), not \(n\), in the formula for the \(n\)th term.

Correct answer: B. 11th term

Explanation: Here, the first term is \(a=3\) and the common difference is \(d=8\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(3+(n-1)\times8=83\), giving \((n-1)\times8=80\), so \(n=11\). Therefore, 83 is the eleventh term. The twelfth term would be \(91\), so that option is not correct. Exam tip: To find the position of a number in an AP, equate it to \(a_n\) and solve for \(n\).

Correct answer: C. -11

Explanation: The nth term of an arithmetic progression is given by \(a_n=a+(n-1)d\). Therefore, \(a_8=17+(8-1)(-4)=17-28=-11\). Hence, option C is correct. A value such as \(-9\) can result from using the wrong term number or forgetting the \((n-1)\) factor. Exam tip: when \(d\) is negative, each successive term decreases.

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

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 end terms. \(q^2=pr\) is associated with a GP. Exam tip: use equal differences to test an AP quickly.

Correct answer: C. 648

Explanation: Here, the first term is \(a=2\), the common difference is \(d=4\), and \(n=18\). The 18th term is \(a_{18}=a+17d=2+17\times4=70\). Therefore, \(S_{18}=\frac{18}{2}(a+a_{18})=9(2+70)=648\). Hence, 648 is correct. An option such as 638 can result from a small error in finding the last term or multiplication. Exam tip: for the \(n\)th term, use \((n-1)d\), not \(nd\).

Correct answer: C. (7)

Explanation: The direct answer is option C, 7. In an AP, moving from the fifth term to the ninth term involves 9−5=4 equal differences. The value increases from 28 to 56, so the total increase is 56−28=28. Therefore 4d=28 and d=28÷4=7. Option A, 5, would produce only 4×5=20 across the four gaps, not 28. Option B, 6, would produce 24, so it is too small. Option C, 7, produces 4×7=28 and is correct. Option D, 8, would produce 32, which is too large. Notice that the first term is unnecessary: subtracting a5 from a9 cancels it and leaves four copies of d. Memory cue: for two known AP terms, use d=(later term−earlier term)/(later position−earlier position).

Correct answer: A. The common difference is \(0\).

Explanation: For an AP, \(a_q-a_p=(q-p)d\). Since \(a_p=a_q\) and \(p\ne q\), we get \((q-p)d=0\), so \(d=0\); hence every term is equal. Exam tip: equal terms at different positions indicate a zero common difference.

Correct answer: B. (77)

Explanation: The position difference is (11) and (d=7), so the difference is (11\times7=77). In exams, use ((m-n)d) directly for the difference of two terms.

Correct answer: B. (252)

Explanation: The first seven terms are (12,20,28,36,44,52,60), and their sum is (252). In exams, you can verify the sum by listing terms from the rule.

Correct answer: C. Eleventh term

Explanation: Here, the first term is \(a=19\) and the common difference is \(d=26-19=7\). The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Thus, \(89=19+(n-1)\times7\), so \(70=7(n-1)\), giving \(n=11\). Therefore, 89 is the eleventh term. The tenth term is \(19+9\times7=82\), so it is not correct. Exam tip: identify \(a\) and \(d\) first, then equate the given value to \(a_n\).

Correct answer: C. 8

Explanation: In an arithmetic progression, the second term is
\(a_2=a+d=14\) and the eighth term is
\(a_8=a+7d=50\). Subtracting the equations gives
\(6d=36\), so
\(d=6\). Hence,
\(a=14-6=8\). Therefore, the first term is 8. Option 6 is the common difference, not the first term. Exam tip: when two terms are given, subtract their equations to find
\(d\) quickly.

Correct answer: C. 10th term

Explanation: Here, the first term is \(a=45\) and the common difference is \(d=-5\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(45+(n-1)(-5)=0\) gives \(n-1=9\), so \(n=10\). The ninth term is \(5\), not zero. Exam tip: To find the position of a specified term, write the formula for \(a_n\) and equate it to the given value.

Correct answer: C. \(6\)

Explanation: The formula for an arithmetic progression is \(a_n=a+(n-1)d\). Substituting \(a=7\), \(n=11\), and \(a_{11}=67\) gives \(67=7+10d\). Hence, \(10d=60\), so \(d=6\). If \(d=5\), the eleventh term would be \(57\), so it is not correct. Exam tip: the coefficient of \(d\) in the \(n\)th-term formula is always \(n-1\).

Correct answer: B. 791

Explanation: Here, the first term is \(a=11\), the common difference is \(d=18-11=7\), and \(n=14\). The sum of the first \(n\) terms is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Thus, \(S_{14}=\frac{14}{2}[2(11)+13(7)]=7(22+91)=7\times113=791\). Therefore, the correct answer is 791. A value such as 773 usually results from an error in using \((n-1)d\) or in multiplication. Exam tip: write down \(a\), \(d\), and \(n\) separately before applying the sum formula.

Correct answer: C. 12

Explanation: Here, the first term is \(a=6\), the common difference is \(d=4\), and the last term is \(l=50\). Using \(l=a+(n-1)d\), we get \(50=6+(n-1)4\), so \(n-1=11\) and \(n=12\). Therefore, there are 12 terms up to 50. With 11 terms, the last term would be 46, not 50. Exam tip: To find the number of terms, equate the last term to \(a+(n-1)d\).

Correct answer: B. 66

Explanation: Here, \(S_4\) denotes the sum of the first four terms: \(S_4=6+13+20+27=66\). Therefore, the correct answer is 66. A value such as 70 can result from misreading the last term or making an addition error. Exam tip: for a small number of terms, add directly; alternatively, use \(S_n=\frac{n}{2}[2a+(n-1)d]\).

Correct answer: C. 16th term

Explanation: For this AP, the first term is \(a=15\) and the common difference is \(d=6\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(105=15+(n-1)\times6\), so \(n-1=15\) and \(n=16\). Therefore, 105 is the 16th term. The 15th term is \(15+14\times6=99\), so it is not correct. Exam tip: do not forget the \(n-1\) in the AP term formula.

Correct answer: B. \(7-4n\)

Explanation: For option B, \(a_{n+1}-a_n=[7-4(n+1)]-(7-4n)=-4\), which is constant, so it is an AP. The differences for \(n^2-3\) change. Exam tip: every linear form \(pn+q\) represents an AP.

Correct answer: C. (73)

Explanation: There are (6) differences from the fifth to the eleventh term, so (a_{11}=31+6\times7=73). In exams, count the difference in positions.

Correct answer: B. 10th term

Explanation: Here, the first term is 90 and the common difference is -9. The general term is \(a_n=90+(n-1)(-9)\). For the term 9, \(90-9(n-1)=9\), so \(n-1=9\) and \(n=10\). Therefore, 9 is the 10th term. The 9th term is 18, so that option is not correct. Exam tip: always use a negative common difference for a decreasing arithmetic progression.

Correct answer: B. (6)

Explanation: From (161=\frac{7}{2}[10+6d]), (d=6). In exams, keep (2a) and ((n-1)d) correct in the (S_n) formula.

Correct answer: C. 132

Explanation: Here, the first term is 16 and the common difference is 6. Thus, \(a_8=16+(8-1)\times6=58\) and \(a_{12}=16+(12-1)\times6=82\). Therefore, \(a_8+a_{12}=58+82=140\), which is not among the given options. The question or options contain an error, so no option is correct. Exam tip: In \(a_n=a+(n-1)d\), be sure to use \(n-1\).

Correct answer: B. 13th term

Explanation: Given \(a_n=62-4n\). For the term whose value is 10, set \(62-4n=10\). Thus, \(4n=52\), so \(n=13\). Hence, the 13th term is equal to 10. The 12th term is \(62-4(12)=14\), so it is not correct. Exam tip: To find a term number, equate the general term \(a_n\) to the given value and solve for \(n\).

Correct answer: B. (396)

Explanation: This is the sum of the first (11) multiples of (6), which is (396). In exams, identify sequences of multiples quickly.