Finding the $n$th term of an AP

Rewrite \(5-2n\) as \(3+(n-1)(-2)\), which matches \(a+(n-1)d\). Thus \(a=3\) and \(d=-2\). Option B wrongly treats the constant 5 as the first term. Exam tip: compare with the standard form first.

The first term is 104 and the common difference is −9, so the nth term is 104 − 9(n − 1). For n = 18, the term is −49; the next term, for n = 19, is −58. Therefore, −49 is the last term greater than −50.

Putting \(n=1\), we get \(a_1=7-3=4\), so the first term is 4. The coefficient of \(n\) gives common difference \(-3\). Option A incorrectly treats 7 as the first term. Exam tip: find \(a_1\) first, then check the coefficient of \(n\).

Answer: 20/3, so option A is correct. The kth term of an arithmetic progression is a_k = a₁ + (k - 1)d. Here the index is k = 3n + 2, the first term is 8, and the common difference is 9. Substitute these values: 197 = 8 + [(3n + 2) - 1]9 = 8 + (3n + 1)9. Expanding gives 197 = 8 + 27n + 9 = 17 + 27n. Therefore 27n = 180, and n = 180/27 = 20/3. Option B, n=7, would make the index 23 and the term 8 + 22(9) = 206. Option C gives index 26 and term 233, while option D gives index 29 and term 260. None equals 197. The important point is subtracting 1 from the index because the first term requires zero jumps; this is a frequent AP error.

In an AP, the difference between two terms equals the difference of their indices multiplied by the common difference. Thus, \(a_{5n}-a_n=(5n-n)d=4nd\). Hence \(205-49=4n\times13\), so \(156=52n\), giving \(n=3\). If \(n=4\), the difference would be \(208\), not 156. Exam tip: In such questions, subtract the given terms instead of first finding the initial term.

The nth term of an AP has the linear form \(a_n=pn+q\), where \(p\) is the common difference. In option A, consecutive terms differ by \(5\). In option B, the differences are not constant. Exam tip: identify the linear form first.

The difference between consecutive terms is \(7\), so the first term is \(a=t-8\) and the common difference is \(d=7\). The \(25\)th term is \(a_{25}=a+24d\). Thus, \(230=(t-8)+24\times7=t+160\), giving \(t=70\). If \(t=66\), the \(25\)th term would be \(226\), so it is not correct. Exam tip: in the formula for the \(n\)th term, multiply \(d\) by \(n-1\), not by \(n\).

Given \(a_n=11n-5\), we have \(a_{4k}=11(4k)-5=44k-5\) and \(a_k=11k-5\). Therefore, \(a_{4k}-a_k=(44k-5)-(11k-5)=33k\). From \(33k=198\), we get \(k=6\). For example, if \(k=5\), the difference would be only 165, not 198. Exam tip: Substitute each index separately in the nth-term formula before subtracting.

The nth term of an AP is \(a_n=a_1+(n-1)d\). Substituting \(a_1=11\) and \(d=-4\) gives option A. In option B, putting \(n=1\) gives 7, not 11. Exam tip: always test \(n=1\) to verify the first term.

Using a_7 = 17(7) + c = 145 gives c = 26. Then a_{3r} = 17(3r) + 26 = 757, so 51r = 731 and r = 731/51 = 43/3. Indeed, 3r = 43, and a_43 = 17(43) + 26 = 757.

The sequence of positive multiples is \(23, 46, 69, \ldots\), and its \(n\)th term is \(23n\). For the last term, \(23n<2000\). Since \(2000\div23\approx86.95\), the greatest integer value of \(n\) is 86. Thus, the last term is \(23\times86=1978\). Although \(2001=23\times87\), it is greater than 2000. Exam tip: for “less than” questions, check that the next multiple does not cross the given limit.

The first term is \(a=703\) and the common difference is \(d=19\), since each successive multiple increases by 19. The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(a_{24}=703+(24-1)\times19=703+437=1140\). Therefore, 1140 is correct. The value 1159 is the next, or 25th, term. Exam tip: do not forget to use \(n-1\) in the nth-term formula.

For an AP, \(a_n=a+(n-1)d\). Thus, \(a_6+a_{18}=2a+22d=156\) and \(a_{10}+a_{22}=2a+30d=252\). Subtracting the first equation from the second gives \(8d=96\), so \(d=12\). Now \(2a+22(12)=156\) gives \(a=-54\). Hence, \(a_{34}=a+33d=-54+33(12)=342\). The value \(324\) does not satisfy the two given conditions. Exam tip: subtract two such sum equations to eliminate \(a\) and find \(d\) quickly.

In an AP, the difference of consecutive terms is the common difference \(d\), so \(a_{n+1}-a_n=d\) must be constant. A constant ratio instead identifies a GP. Exam tip: subtract two consecutive nth-term expressions to test for an AP.

Here, the first term is \(a=35\) and the common difference is \(d=11\). The \(n\)th term is \(a_n=35+11(n-1)\). Using \(35+11(n-1)<600\), we get \(n<52.36\ldots\), so the greatest possible integer value is \(n=52\). Check: \(a_{52}=596<600\), whereas \(a_{53}=607>600\). Therefore, 52 terms are less than 600. Exam tip: For a ‘less than’ condition, verify the next term to confirm the count.

Putting \(n=1\) gives \(a_1=7-3=4\). In \(a_n=a+(n-1)d\), the coefficient of \(n\) is the common difference, so \(d=-3\). Exam tip: always verify the first term by substituting \(n=1\).

In an AP, the common difference is the same between consecutive terms. Here, \(a_9=a_4+5d\) and \(a_{14}=a_4+10d\). Therefore, \(20+5d+20+10d=160\), so \(15d=120\) and \(d=8\). Now, \(a_{28}=a_4+24d=20+24\times8=212\). Hence, 212 is correct. The value 204 does not result when the correct common difference \(d=8\) is used. Exam tip: when one term \(a_r\) is known, use \(a_n=a_r+(n-r)d\) directly.

Here, the first term is \(a=-38\) and the common difference is \(d=13\). Thus, \(a_n=-38+13(n-1)=13n-51\). For \(a_n>300\), we get \(13n-51>300\), so \(n>27\). The least integer value is \(n=28\), and \(a_{28}=313\). Note that 300 is not itself a term of the AP, so 313 is the correct answer. Exam tip: for “greater than,” use \(>\) and then take the least possible integer value of \(n\).

Given \(a_n=6n+q\), we get \(a_{5n}=6(5n)+q=30n+q\). Hence, \(a_{5n}-a_n=(30n+q)-(6n+q)=24n\). Now \(24n=216\), so \(n=216/24=9\). Therefore, the correct option is \(9\). The term \(q\) cancels because it occurs in both terms. Exam tip: In such questions, first write \(a_{5n}\) explicitly and then subtract \(a_n\).

The indices 7, 10 and 13 are equally spaced by 3, so their AP terms are also equally spaced. Therefore, the middle term is the average: \(a_{10}=\frac{a_7+a_{13}}{2}\). Exam tip: look for equally spaced term numbers.

Use the AP relation between term differences and index differences. Since a₁₉ − a₇ = 122 − 38 = 84 and 19 − 7 = 12, we have 12d = 84, giving d = 7. The index of the required term is 7r, so its distance from the seventh term is 7r − 7. Therefore 206 = 38 + (7r − 7) × 7. Simplifying, 168 = 49r − 49, so 217 = 49r and r = 217/49 = 31/7. Hence option C is correct. Verification is direct: when r = 31/7, the index 7r equals 31, and a₃₁ = 38 + (31 − 7) × 7 = 38 + 168 = 206. The integer distractors give different indices and do not satisfy the condition.

The governing formula is a_m = a₁ + (m − 1)d for an arithmetic progression. Here m = 3n + 4, so the multiplier of d is (3n + 4) − 1 = 3n + 3, not 3n + 4. Substitute the values: 275 = 11 + (3n + 3) × 8 = 11 + 24n + 24 = 24n + 35. Hence 24n = 240 and n = 10. Therefore option B is correct. A direct check confirms the result: n = 10 gives m = 3(10) + 4 = 34, and a₃₄ = 11 + (34 − 1) × 8 = 11 + 264 = 275. The other options produce different indices and fail the original equation. Forgetting the minus one is the most likely source of an incorrect answer.

Use the stated linear formula and the difference condition. From aₙ = kn + 17, we obtain a₂₀ = 20k + 17 and a₇ = 7k + 17. Subtracting gives a₂₀ − a₇ = (20k + 17) − (7k + 17) = 13k, because the constant terms cancel. Since the difference is 104, 13k = 104 and hence k = 8. The required term is therefore a₃₁ = 31k + 17 = 31 × 8 + 17 = 248 + 17 = 265. Thus option B is correct. Options A, C and D can arise from an incorrect difference, a wrong value of k, or an arithmetic error. The formula is linear in n, consistent with an arithmetic progression.

The governing concept is the nth term of an arithmetic progression together with an inequality. The first term is a = 33 and the common difference is d = 48 − 33 = 15. Therefore aₙ = 33 + 15(n − 1). We require the least term greater than 1000, so 33 + 15(n − 1) > 1000. This gives 15(n − 1) > 967 and n − 1 > 64.466..., so the least integer value is n − 1 = 65, or n = 66. The term is a₆₆ = 33 + 65 × 15 = 33 + 975 = 1008. The preceding term is a₆₅ = 993, which is below 1000, confirming that 1008 is the first qualifying term. Hence option B is correct.