Arithmetic Progression

The governing concept is the explicit formula for an arithmetic sequence, aₙ = a₁ + (n − 1)d. From the first term to the fourth term there are three equal differences, not four. Therefore, 3d = a₄ − a₁ = 20 − 8 = 12, so d = 4. Substituting a₁ = 8 and d = 4 gives aₙ = 8 + 4(n − 1) = 8 + 4n − 4 = 4n + 4. Hence option B is correct. A gives a₁ = 12, which contradicts the data. C gives a₁ = 8 but a₄ = 23, and D gives a₁ = 8 but a₄ = 17. The essential checking method is to evaluate any proposed rule at n = 1 and n = 4; only B produces both given terms.

In an AP, the difference between consecutive terms is a constant d; for example, 3, 7, 11 has common difference 4. Doubling terms indicates a geometric pattern, not an AP. Exam tip: subtract consecutive terms to test an AP.

The governing concept is the nth-term formula for an arithmetic progression: aₙ = a + (n − 1)d. In the sequence 42, 36, 30, 24, each term decreases by 6, so the first term is a = 42 and the common difference is d = −6. For n = 8, a₈ = 42 + (8 − 1)(−6) = 42 + 7(−6) = 42 − 42 = 0. Therefore option D is correct. A direct continuation also confirms the result: the fifth, sixth, seventh, and eighth terms are 18, 12, 6, and 0. Option A is only the common difference, option B reverses its sign, and option C is the sixth term rather than the eighth. The negative difference is essential.

To find which term is 91, put \(a_n=91\) in the general term: \(4n+3=91\). Thus, \(4n=88\), so \(n=22\). Therefore, 91 is the 22nd term. The 21st term is \(4(21)+3=87\), so it is not correct. Exam tip: To find the term number of a given value, equate \(a_n\) to that value and solve for \(n\).

For an arithmetic progression, \(a_n=a+(n-1)d\). Here, \(a=14\) and \(a_{15}=84\), so \(84=14+(15-1)d=14+14d\). Thus, \(70=14d\), giving \(d=5\). If the common difference were 4, the fifteenth term would be \(14+14\times4=70\), not 84. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

The middle term is (x=\frac{6+18}{2}=12). In exams, the middle term of three AP terms is the average.

In an arithmetic progression, the difference between consecutive terms is constant. For \(a_n=7-3n\), \(a_{n+1}-a_n=[7-3(n+1)]-(7-3n)=-3\), which is fixed. For \(n^2-3\), the difference changes. Exam tip: test \(a_{n+1}-a_n\).

The rule (a_n=29-4n) gives (25) at (n=1) and (21) at (n=2). In exams, match the first term in a decreasing progression.

The formula for an arithmetic progression is \(a_n=a+(n-1)d\). Substituting \(a_6=43\), \(a=8\), and \(n=6\) gives \(43=8+5d\). Thus, \(5d=35\), so \(d=7\). If 8 were chosen, the sixth term would be \(8+5\times8=48\), not 43. Exam tip: use \(n-1\), not \(n\), in the formula for \(a_n\).

Here, the first term is \(a=4\) and the common difference is \(d=6\). The \(n\)th term is \(a_n=a+(n-1)d\). So, \(4+(n-1)6=70\) gives \(6(n-1)=66\), hence \(n=12\). Therefore, 70 is the twelfth term. The eleventh term is \(64\), so it is not correct. Exam tip: when asked for a term number, equate \(a_n\) to the given term and solve for \(n\).

The formula for the nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Thus, \(a_9=13+(9-1)(-3)=13-24=-11\). Therefore, \(-11\) is correct. \(-9\) can result from using an incorrect number of common differences. Exam tip: when \(d\) is negative, check the sign after multiplication carefully.

If \(T_p=T_q\) and \(p\ne q\), then \(a+(p-1)d=a+(q-1)d\), giving \((p-q)d=0\). Hence \(d=0\), so every term is equal. Exam tip: note the condition that the positions are distinct.

Here, the first term is \(a=1\), the common difference is \(d=3\), and \(n=20\). The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Therefore, \(S_{20}=\frac{20}{2}[2(1)+19(3)]=10(59)=590\). Hence, 590 is correct. A result such as 600 may come from using an incorrect number of common differences. Exam tip: to reach the \(n\)th term from the first term, use \(n-1\) common differences.

The direct answer is option C, 6. In an AP, the difference between two terms equals the number of equal gaps between their positions multiplied by d. From the fourth term to the eighth term there are 8−4=4 gaps. The term difference is a8−a4=44−20=24. Hence 4d=24, so d=24÷4=6. Option A, 4, would give a difference of only 16 across four gaps. Option B, 5, would give 20, not 24. Option C, 6, gives 4×6=24 and is correct. Option D, 7, would give 28, which is too large. There is no need to find the first term because subtracting the given terms removes it: (a+7d)−(a+3d)=4d. Memory cue: divide the difference in values by the difference in positions.

In an arithmetic progression, \(a_n=a_1+(n-1)d\). Therefore, \(a_5=a_1+4d\). Substituting the given values, \(27=a_1+4\times5=a_1+20\), so \(a_1=7\). If \(a_1=9\), the fifth term would be \(29\), not \(27\). Exam tip: there are \(n-1\) common differences between the first term and the \(n\)th term.

The position difference is (10) and (d=7), so the difference is (10\times7=70). In exams, use ((m-n)d) directly for the difference of two terms.

The first six terms are (7,16,25,34,43,52), and their sum is (186). In exams, you can verify the sum by listing terms from the rule.

Here, the first term is \(a=17\) and the common difference is \(d=24-17=7\). The \(n\)th term is \(a+(n-1)d\). Thus, \(17+(n-1)\times7=80\), so \((n-1)\times7=63\) and hence \(n=10\). Therefore, 80 is the 10th term. The 9th term is 73, so it is not correct. Exam tip: To find a term number, equate the given value directly to \(a+(n-1)d\).

For an arithmetic progression, \(a_n=a+(n-1)d\). Thus, \(a_2=a+d=12\) and \(a_6=a+5d=32\). Subtracting the equations gives \(4d=20\), so \(d=5\). Substituting \(d=5\) into \(a+d=12\) gives \(a=7\). Therefore, the correct answer is 7. Option 6 may seem close, but it does not satisfy the first-term calculation from \(a+d=12\). Exam tip: when two terms are given, subtract their equations first to find \(d\).

Here, the first term is \(a=30\) and the common difference is \(d=-3\). The \(n\)th term is \(a_n=a+(n-1)d\). For the term to be zero, \(30+(n-1)(-3)=0\). This gives \(30-3n+3=0\), so \(n=11\). Hence, the eleventh term is zero. The tenth term is \(3\), not zero. Exam tip: To find a term number in an AP, substitute the given term value in \(a_n\) and solve for \(n\).

For an arithmetic progression, \(a_n=a+(n-1)d\). Hence, \(a_9=2+(9-1)d=2+8d\). Given \(a_9=50\), we get \(50=2+8d\), so \(8d=48\) and \(d=6\). If \(d=5\), the ninth term would be \(2+8\times5=42\), not 50. Exam tip: in the \(n\)th-term formula, the coefficient of \(d\) is always \(n-1\).

Here, the first term is \(a=9\), the common difference is \(d=16-9=7\), and \(n=12\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_{12}=\frac{12}{2}[2(9)+11(7)]=6(18+77)=570\). Therefore, 570 is correct. A value such as 558 may result from using the wrong number of terms or misusing \((n-1)\). Exam tip: write down \(a\), \(d\), and \(n\) before substituting in the sum formula.

Here, the first term is 5 and the common difference is 4. Using \(a_n=a+(n-1)d\), we get \(41=5+(n-1)\times4\). Thus, \(n-1=9\), so \(n=10\). Option 9 is incorrect because 37 is the ninth term. Exam tip: To find the number of terms up to a given last term, equate that term to \(a_n\) and solve for \(n\).

Here, \(S_4\) denotes the sum of the first four terms. Thus, \(S_4=3+8+13+18=42\). Therefore, 42 is correct. Getting 40 usually indicates an addition error in one of the terms. Exam tip: for a few terms, add directly; alternatively, use \(S_n=\frac{n}{2}[2a+(n-1)d]\).

Here, the first term is \(a=12\) and the common difference is \(d=18-12=6\). Using \(a_n=a+(n-1)d\), we get \(12+(n-1)\times6=96\). Thus, \((n-1)\times6=84\), so \(n-1=14\) and \(n=15\). Hence, 96 is the 15th term. The 14th term is \(90\), so that nearby option is incorrect. Exam tip: identify \(a\) and \(d\) first, then substitute them in \(a_n=a+(n-1)d\).