Finding the $n$th term of an AP

Correct answer: A. First term \(3\), common difference \(-2\)

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.

Correct answer: A. −49

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.

Correct answer: B. First term 4 and common difference \(-3\)

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 \(n\).

Correct answer: A. 20/3

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.

Correct answer: A. 3

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.

Correct answer: A. \(a_n=5n-2\)

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.

Correct answer: C. 70

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 \(n\).

Correct answer: C. 6

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.

Correct answer: A. \(a_n=11-4(n-1)\)

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.

Correct answer: C. 43/3

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.

Correct answer: B. 1978

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.

Correct answer: B. 1140

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.

Correct answer: D. 342

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.

Correct answer: A. \(a_{n+1}-a_n\) is the same constant for every positive integer \(n\)

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.

Correct answer: C. 52

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.

Correct answer: A. First term 4 and common difference -3

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 \(n=1\).

Correct answer: C. 212

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.

Correct answer: B. 313

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 \(n\).

Correct answer: C. \(9\)

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 \(a_n\).

Correct answer: A. \(a_{10}=\frac{a_7+a_{13}}{2}\)

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.

Correct answer: C. 31/7

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.

Correct answer: B. 10

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.

Correct answer: B. 265

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.

Correct answer: B. 1008

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.