Finding the $n$th term of an AP

The direct answer is option C: 289. In an AP, terms change by the same common difference. The indices 18 and 46 have midpoint (18 + 46)/2 = 32, so a32 lies exactly halfway between a18 and a46. In an AP, equally spaced terms have values equally spaced too. Therefore a32 = (a18 + a46)/2 = (135 + 443)/2 = 578/2 = 289. Option C is correct. Option A, 279, results from an incorrect average or subtraction. Option B, 285, does not equal the required midpoint. Option D, 295, is also not the average of 135 and 443. We can verify using differences: from index 18 to 32 there are 14 steps, and from 32 to 46 there are also 14 steps, so the middle term must be the numerical midpoint. Exam cue: when the index is exactly halfway between two AP indices, average the two corresponding terms.

The difference between consecutive terms is \((v-3)-(v-12)=9\), so \(d=9\) and the first term is \(a=v-12\). The nth term is \(a_n=a+(n-1)d\). Hence, \(438=(v-12)+30\times9=v+258\), giving \(v=180\). If \(v=176\), the 31st term would be \(434\), so it is not correct. Exam tip: In an AP, multiply the common difference by \(n-1\), not by \(n\).

Given \(a_n=15n-8\), we get \(a_{6k}=15(6k)-8=90k-8\) and \(a_{2k}=15(2k)-8=30k-8\). Therefore, \(a_{6k}-a_{2k}=(90k-8)-(30k-8)=60k\). Using \(60k=300\), we obtain \(k=5\). If \(k=4\), the difference would be only \(240\), so it is not correct. Exam tip: when subtracting two terms of this form, the identical constant terms cancel out.

In \(a_n=a+(n-1)d\), the coefficient of \(n\) is the common difference \(d\). Here it is \(-4\), so \(d=-4\) and the terms decrease. As a quick check, putting \(n=1\) gives the first term \(3\), not \(7\).

The sequence of positive multiples is \(37,74,111,\ldots\), whose \(n\)th term is \(37n\). For the last term, \(37n<3600\). Now \(37\times97=3589\), whereas \(37\times98=3626>3600\). Hence, 3589 is the greatest multiple of 37 less than 3600. Although 3626 is the next multiple, it exceeds the given limit. Exam tip: divide by the common difference, take the integer part, and multiply back to find the required multiple.

The first term of the AP is \(a=1209\), and the common difference is \(d=1240-1209=31\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{29}=1209+(29-1)\times31=1209+868=2077\). Hence, 2077 is correct. Using \(29\times31\) would incorrectly add 29 gaps; from the first term to the 29th term, there are only 28 gaps. Exam tip: always use \((n-1)\) in the nth-term formula.

For an AP, \(a_n=a+(n-1)d\). Hence, \(a_8+a_{24}=2a+30d=320\) and \(a_{14}+a_{30}=2a+42d=512\). Subtracting the equations gives \(12d=192\), so \(d=16\). Then \(2a+30(16)=320\), giving \(a=-80\). Therefore, \(a_{46}=a+45d=-80+45(16)=640\). A common mistake is to use \(46d\) instead of \((46-1)d\) in the term formula. Exam tip: when sums of AP terms are given, form both linear equations and subtract them first.

First, a₁₃ = 4(36) − 12 = 132. The terms a₇ and a₁₃ are six positions apart, so 6d = 132 − 36 = 96 and d = 16. Since a₃₇ is 30 positions after a₇, a₃₇ = 36 + 30(16) = 516.

Here, the first term is 56 and the common difference is 17. Therefore, \(a_n=56+17(n-1)\). Using \(a_n<1500\), we get \(56+17(n-1)<1500\), so \(n-1<84.94\ldots\). Hence, the greatest integer value of n is 85. In fact, \(a_{85}=1484\), whereas \(a_{86}=1501\), which is not less than 1500. Exam tip: For “less than,” do not include a term equal to the limiting value.

The nth term of an AP is \(a_n=a+(n-1)d\), so it must be linear in n. For option B, \(a_{n+1}-a_n=2n-2\), which changes with n; hence the common difference is not constant. Exam tip: check whether the first difference is fixed.

Using aₙ = a₆ + (n − 6)d, we get a₁₄ + a₂₂ = (42 + 8d) + (42 + 16d) = 84 + 24d = 420, so d = 14. Therefore, a₃₈ = 42 + 32(14) = 490.

Here, the first term is \(a=-68\) and the common difference is \(d=19\). Thus, \(a_n=-68+19(n-1)\). For \(a_n>700\), \(-68+19(n-1)>700\), giving \(n>41.42\). The smallest integer value is \(n=42\). Therefore, \(a_{42}=711\), so 711 is the first term greater than 700. Although 692 is the nearest preceding term, it is less than 700. Exam tip: for the “first term greater than” condition, solve the inequality and take the smallest valid integer \(n\).

Given \(a_n=10n+q\), we get \(a_{7n}=70n+q\) and \(a_{3n}=30n+q\). Hence, \(a_{7n}-a_{3n}=(70n+q)-(30n+q)=40n\). Therefore, \(40n=520\), so \(n=13\). The constant \(q\) cancels because it occurs in both terms. Exam tip: for terms such as \(a_{7n}\) and \(a_{3n}\), directly substitute \(7n\) and \(3n\) respectively in the term formula.

Comparing with \(a_n=a+(n-1)d\), we get \(a=4\) and \(d=-7\). A negative common difference means each next term is 7 less, so the AP decreases. Exam tip: the coefficient of \((n-1)\) is the common difference.

The common difference is d = (205 − 77) ÷ (29 − 13) = 8. From a₂₉ to a₄₇ there are 18 steps, so a₄₇ = 205 + 18 × 8 = 349.

The direct answer is option B: p = 4. For an AP, the difference between terms whose indices differ by k is kd. The indices 4p and p differ by 4p − p = 3p. Therefore a4p − ap = 3pd. Substitute the given values: 188 − 56 = 132, so 3p × 11 = 132. Hence 33p = 132, and dividing both sides by 33 gives p = 4. Option B is correct. Option A, 3, would give 33 × 3 = 99, not 132. Option C, 5, would give 165, and option D, 6, would give 198; neither matches the given term difference. The calculation assumes the stated AP difference d = 11. Exam cue: subtract the term values, subtract their indices, then use value difference = index gap × common difference.

Expanding option A gives \(a_n=n^2+5n+4-n^2=5n+4\). It has the form \(an+b\), so its common difference is the constant \(5\). Option B contains an \(n^2\) term, so successive differences cannot remain constant. Exam tip: an AP’s nth term is linear in \(n\).

In this AP, the first term is 260 and the common difference is \(d=-15\). \((-100)\) is itself a term of the AP, since \(260-15\times 24=-100\). As each successive term decreases by 15, the term after \((-100)\) is \((-100)-15=-115\). Therefore, the first term less than \((-100)\) is \((-115)\). Although \((-130)\) is also less than \((-100)\), it occurs later. Exam tip: when asked for the “first” term beyond a boundary, check the term immediately after the boundary term.

In an AP, \(a_{n+1}-a_n\) must be constant for every \(n\). For option A, \([5(n+1)-2]-(5n-2)=5\). In B, the difference changes with \(n\). Exam tip: look for the linear form \(pn+q\).

The governing concept is the nth-term relation in an arithmetic progression: aₙ = aₘ + (n − m)d, where d is the common difference. Since a₇ = 22, the given condition gives a₁₆ = 4(22) + 2 = 90. The indices 16 and 7 differ by 9, so a₁₆ − a₇ = 9d. Hence 90 − 22 = 9d, which gives d = 68/9. To reach the 34th term from the 7th term, move through 34 − 7 = 27 equal steps. Thus a₃₄ = 22 + 27(68/9) = 22 + 204 = 226. Therefore option A is correct. Options B, C and D arise from incorrect multiplication or from using the wrong index difference; none satisfies the AP relation with the stated data.

The \(n\)th term of an AP has the linear form \(an+b\), so consecutive differences must be constant. For \(n^2+1\), the differences are \(3,5,7,\ldots\), not constant. Exam tip: identify whether the formula is linear in \(n\).

Using a₉ = 128 gives 11 × 9 + c = 128, so c = 29. Then a₄ᵣ = 44r + 29 = 381, so 44r = 352 and r = 8.

In an AP, the difference between consecutive terms is constant. Here, \(a_{n+1}-a_n=[4(n+1)-9]-(4n-9)=4\), so it is an AP. A constant ratio identifies a GP instead. Exam tip: check the coefficient of \(n\) for the common difference.

Using \(a_n=a+(n-1)d\), \(a_7+a_{19}=(a+6d)+(a+18d)=2a_{13}\). So option A is always true; B has the wrong middle term. Exam tip: pair terms equidistant from the centre.

The nth term of an AP is \(a_n=a+(n-1)d\), so it is linear in \(n\). In \(n^2+2\), the consecutive differences are \(3,5,7\dots\), not constant. Exam tip: check whether the first differences remain equal.