Arithmetic Progression

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.

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.

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 \(n\).

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.

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.

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 \(nd\).

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).

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.

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.

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.

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 \(a_n\).

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.

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.

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 \(n-1\).

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.

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 \(a+(n-1)d\).

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 \(S_n=\frac{n}{2}[2a+(n-1)d]\).

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.

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.

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.

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.

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

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 \(n-1\).

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 \(n\).

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