Finding the $n$th term of an AP

There are 19 − 7 = 12 common differences between a₇ and a₁₉, so d = (103 − 31) ÷ 12 = 6. From a₁₉ to a₃₄ there are 34 − 19 = 15 steps; hence a₃₄ = 103 + 15 × 6 = 193.

Use the AP relation from a known term: \(a_n=a_r+(n-r)d\). Thus, \(a_{25}=a_{10}+(25-10)d=43+15\times5=118\). Hence, option B is correct. The condition \(a_4+a_{16}=86\) confirms \(2a_{10}=86\), since the 4th and 16th terms are equally placed around the 10th term; it is not needed separately to find \(a_{25}\). Exam tip: use the index difference \(25-10=15\), not \(25-1\).

The direct term rule is aₙ = 3n + p, but the unknown constant p must first be found from the given term. Put n = 12 and use a₁₂ = 50: 50 = 3(12) + p = 36 + p, so p = 14. Now substitute n = 27 into the same rule: a₂₇ = 3(27) + 14 = 81 + 14 = 95. Therefore option B is correct. The expression also represents an arithmetic progression because increasing n by one increases aₙ by 3, so its common difference is 3. As a second check, a₂₇ − a₁₂ = 15 × 3 = 45, and 50 + 45 = 95. Option A may omit or mishandle p, while options C and D arise from arithmetic or index errors.

The index difference between a₂ₘ and aₘ is 2m − m = m, so a₂ₘ − aₘ = md. Thus 58 − 31 = 3m, giving 27 = 3m and m = 9.

Here, the first term is \(a=41\) and the common difference is \(d=35-41=-6\). Therefore, \(a_n=41+(n-1)(-6)=47-6n\). From \(47-6n<-50\), we get \(n>97/6\), so the least possible integer value is \(n=17\). Thus, \(a_{17}=47-6(17)=-55\), making \((-55)\) the correct answer. \((-49)\) is the 16th term, but it is not less than \((-50)\). Exam tip: after solving an inequality for the first required term, choose the least integer greater than the obtained value.

There are 7 positions between the 5th and 12th terms. Thus, \(a_{12}-a_5=7d\), so \((x+42)-(x+7)=35=7d\), giving \(d=5\). From the 12th term to the 20th term, there are 8 positions; hence \(a_{20}=a_{12}+8d=x+42+8(5)=x+82\). The option \(x+77\) results from incorrectly using a gap of only 7 positions. Exam tip: the difference between term numbers gives the multiplier of \(d\).

Given \(a_3=18\), we get \(a_{11}=4a_3=4\times18=72\). In an AP, \(a_{11}-a_3=(11-3)d=8d\). Thus, \(8d=72-18=54\), so \(d=\frac{27}{4}\). Now \(a_{15}=a_{11}+4d=72+4\times\frac{27}{4}=99\). Hence, 99 is correct. The value 96 would result from not accounting correctly for the common difference between the terms. Exam tip: use the difference in term numbers to find \(d\) when two terms are given.

Given \(a_8=27\) and \(a_{18}=77\), the common difference is \(d=\frac{77-27}{18-8}=5\). Using \(a_n=a_8+(n-8)d\), we get \(152=27+(n-8)\times5\). Thus, \(125=5(n-8)\), so \(n-8=25\) and \(n=33\). Hence, 152 is the 33rd term. The 32nd term would be \(147\), so it is not correct. Exam tip: When two terms are given, first find \(d\) using the difference in their term numbers.

Given \(a_n=kn+5\), we have \(a_{15}=15k+5\) and \(a_6=6k+5\). Thus, \(a_{15}-a_6=(15k+5)-(6k+5)=9k=63\), so \(k=7\). Therefore, \(a_{20}=20k+5=20\times7+5=145\). Hence, 145 is the correct option. Choosing 140 would incorrectly omit the constant term \(+5\). Exam tip: when subtracting two terms, the same constant term cancels out.

Here, the first term is \(a=2\) and the common difference is \(d=7\). Thus, the \(n\)th term is \(a_n=2+7(n-1)=7n-5\). From \(7n-5>200\), we get \(n>\frac{205}{7}\), so the smallest integer value is \(n=30\). Therefore, \(a_{30}=205\), making 205 the first term greater than 200. Although 198 is the nearest preceding term, it is not greater than 200. Exam tip: for the first term greater than a given number, solve the inequality and take the smallest valid integer value of \(n\).

Since \(p=6\), we have \(a_p=a_6=17\) and \(a_{p+8}=a_{14}=65\). Thus, \(a_{14}-a_6=8d=65-17=48\), so \(d=6\). Now, \(a_{30}=a_6+(30-6)d=17+24\times6=161\). Hence, 161 is correct. Choosing 155 would not use the correct gap of \(30-6=24\) terms. Exam tip: In questions with indexed terms, substitute the value of \(p\) first and write the actual term numbers.

In an AP, the sum of terms equidistant from a middle term equals twice that middle term. Since the index 6 lies midway between 3 and 9, \(a_3+a_9=2a_6\). Thus, \(2a_6=72\), so \(a_6=36\). Values such as 34 or 38 would not give the stated sum of 72. Exam tip: use \(a_{m-r}+a_{m+r}=2a_m\) for symmetric terms in an AP.

The key property is symmetry in an arithmetic progression. Terms equally distant from a central term have a sum equal to twice that central term: a₅₋₃ + a₅ + a₅₊₃ = 3a₅. Here a₂ and a₈ are three positions below and above a₅, respectively. Therefore a₂ + a₈ = 2a₅, and the given equation becomes 2a₅ + a₅ = 111, or 3a₅ = 111. Dividing by 3 gives a₅ = 37. Hence option B is correct. No first term or common difference is needed. The other choices would make the total 105, 117, or 123 under the same symmetry relation.

Using the AP formula, \(a_9=12+(9-1)d\). Thus, \(68=12+8d\), so \(d=7\). Now \(a_{22}=12+(22-1)\times7=12+147=159\). Choosing 166 would result from using an incorrect common difference instead of 7. Exam tip: first find \(d\) from the given term, then use \(n-1\) for the required term.

Here, \(a=84.5\) and \(d=-2.75\). Thus \(a_{16}=a+15d=84.5-15(2.75)=43.25\) cm. The student used 16 instead of \(n-1\). Exam tip: always count the number of jumps from the first term.

For an AP, \(a_n=a_1+(n-1)d\). Thus, \(a_{21}-a_{10}=11d=-55\), so \(d=-5\). Now, using \(a_{10}=a_1+9d\), we get \(0=a_1+9(-5)\), hence \(a_1=45\). If 40 were chosen, the tenth term would be \(-5\), not the given value 0. Exam tip: When two terms are given, subtract them first to find \(d\) quickly.

The nth term of an AP has the linear form \(a_n=a+(n-1)d\). For \(n^2+1\), the first three terms are 2, 5, 10, with differences 3 and 5, so it is not an AP. Exam tip: a linear expression in n represents an AP.

The sequence is an arithmetic progression because the difference between consecutive terms is constant. Here the common difference is \(d = 88 - 97 = -9\), so every next term is 9 less than the preceding term. Since the sequence decreases, we can list terms near 20 or use the nth-term formula \(a_n = 97 + (n-1)(-9)\). We need the first term that is strictly less than 20.

The terms descend as 97, 88, 79, 70, 61, 52, 43, 34, 25, 16, 7, and so on. The term 25 is still greater than 20, while the next term, 16, is less than 20. Therefore the first required term is 16, so option A is correct. The number 18 is not in this progression because its terms change by 9 each time.

In an AP, \(a_{n+1}-a_n=d\) is constant. Hence the difference of consecutive first differences is zero, giving option A. A constant ratio identifies a GP instead. Exam tip: check first differences first.

In an AP, \(a_{13}-a_3=(13-3)d\). Hence \(51-11=10d\), so \(d=4\). Now use \(a_{3r}=a_3+(3r-3)d\): \(91=11+(3r-3)\times4\). Thus \(3r-3=20\), giving \(3r=21\) and \(r=7\). If \(r=6\), the term would be \(a_{18}\), whose value is 71, not 91. Exam tip: first find \(d\) from the two known terms and carefully use the difference between their indices.

Direct answer: Option A, \(11x-8\). In an AP, terms equally far apart have differences proportional to the number of steps. The given terms are at positions 2, 5, and 8, so each pair is separated by three positions. Calculate the difference from the fifth to the eighth term: \(a_8-a_5=(8x-5)-(5x-2)=3x-3\). The 11th term is also three positions after the eighth term, so the same difference must be added: \(a_{11}=a_8+(3x-3)=(8x-5)+(3x-3)=11x-8\). Option A is correct. Option B, \(11x-6\), adds only 2 instead of the required difference and does not follow the given AP. Option C, \(10x-7\), has the wrong coefficient and constant after the correct addition. Option D, \(12x-9\), adds too much and is not the next three-step AP value. Another check is that the expressions change by \(3x-3\): from \(a_2\) to \(a_5\), from \(a_5\) to \(a_8\), and from \(a_8\) to \(a_{11}\). Exam cue: equal gaps in position mean equal changes in an AP.

Here the first term is 14 and the common difference is 9, so the nth term is \(a_n=14+9(n-1)\). On putting \(a_n=347\), we get \(14+9(n-1)=347\), which gives \(n=38\). Thus, 347 is a term of the AP and is less than 350. The next term is \(347+9=356\), which is greater than 350. Therefore, 347 is the greatest required term. Exam tip: After finding a term near an upper limit, always check the next term as well.

In an AP, \(a_{n+2}-a_n=2d\). Thus, \(2d=18\), so \(d=9\). The 25th term is 16 positions after the 9th term; hence \(a_{25}=a_9+16d=74+16\times9=218\). Option 216 is not possible because its difference from 74 is 142, which is not a multiple of 9. Exam tip: subtract the term indices and multiply the result by the common difference.

Let the first term be \(a\) and the common difference be \(d\). Then \(a_1+a_4=a+(a+3d)=2a+3d=38\), while \(a_7=a+6d=44\). Multiplying the second equation by 2 gives \(2a+12d=88\). Subtracting the first equation gives \(9d=50\), so \(d=\frac{50}{9}\). Hence, \(a_{16}=a_7+9d=44+9\times\frac{50}{9}=94\). The value 80 results from incorrectly taking \(d=4\). Exam tip: eliminate the common term by multiplying one AP equation suitably before subtracting.

Given \(a_3=13\), \(a_6=2(13)+5=31\). In an AP, \(a_6-a_3=(6-3)d=3d\). Hence, \(31-13=18=3d\), so \(d=6\). Now \(a_{12}=a_3+(12-3)d=13+9\times6=67\). Therefore, 67 is correct. A value such as 69 can result from incorrectly counting the number of common differences. Exam tip: when using \(a_n=a_r+(n-r)d\), carefully use \(n-r\) as the number of gaps.