Finding the $n$th term of an AP

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 with first term a_1=97 and common difference d=88-97=-9. Its nth term is a_n=97+(n-1)(-9). To identify the first term below 20, inspect the terms near the boundary: 97, 88, 79, 70, 61, 52, 43, 34, 25, 16. The term 25 is still greater than 20, while the next term, 16, is less than 20. Therefore the first term satisfying the condition is 16, so option A is correct. The later value 7 also satisfies the inequality but is not the first such term; 25 fails the condition.

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.

The given terms are equally spaced, so (a_8-a_5=3x-3) is the equal group difference. (a_{11}=a_8+(3x-3)=11x-8).

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.

The first term is \(a=-7\), and the common difference is \(d=\frac{-3}{2}-(-7)=\frac{11}{2}\). Using \(a_n=a+(n-1)d\), we get \(a_{18}=-7+17\times\frac{11}{2}=\frac{-14+187}{2}=\frac{173}{2}\). Hence, \(\frac{173}{2}\) is correct. The option \(\frac{175}{2}\) can result from a small error in counting terms or arithmetic. Exam tip: for an AP involving fractions, write terms with a common denominator first to check the common difference accurately.

Given \(a_8=28\), \(a_{20}=3a_8=3\times28=84\). In an AP, \(a_{20}-a_8=(20-8)d=12d\). Hence, \(84-28=12d\), so \(d=\frac{14}{3}\). Now \(a_{26}=a_{20}+(26-20)d=84+6\times\frac{14}{3}=84+28=112\). Therefore, 112 is correct. Choosing 106 would mean adding the increase over six terms incorrectly. Exam tip: when two distant terms are given, use the difference of their indices to find \(d\).

In an AP, \(a_{15}-a_5=(15-5)d=10d\). Since \(a_{15}=a_5+50\), we get \(10d=50\), so \(d=5\). Now, \(a_{31}=a_5+(31-5)d=24+26\times5=154\). Hence, 154 is correct. Although 158 is close, it does not follow from the given common difference \(d=5\). Exam tip: When comparing two terms, use the difference between their term numbers.

Given \(a_n=an+b\), we have \(a_{14}-a_4=10a\). Thus, \(71-21=10a\), so \(a=5\). From the 14th term to the 30th term, there are 16 term-intervals. Therefore, \(a_{30}=a_{14}+16\times5=71+80=151\). The value 147 would result from using an incorrect interval count. Exam tip: subtract two given terms to eliminate the constant \(b\) and find the common difference/slope directly.

The first term is 64 and the common difference is \(d=57-64=-7\). Thus, each successive term decreases by 7: \(1,-6,-13,-20\). Here, \((-13)\) is greater than \((-20)\), while the next term is \((-20)\), which is not greater than itself. Therefore, the answer is \((-13)\). Exam tip: For “greater than”, do not include the boundary value itself.

From (75=240+(n-1)(-15)), (165=15(n-1)), so (n=12). In word problems, take decrease as negative (d).

The number of questions solved each day forms an AP with first term \(a=18\) and common difference \(d=4\). The number solved on the \(n\)th day is \(a_n=a+(n-1)d\). Thus, \(102=18+(n-1)4\), so \(84=4(n-1)\) and \(n-1=21\). Hence, \(n=22\). On the 21st day, the student solves 98 questions, so 102 questions are solved on the 22nd day. Exam tip: identify whether the question asks for a term of the AP or for the total sum.

From \(123=3+2n\cdot4\), \(120=8n\), so \(n=15\). In index (2n+1), subtract (1) to get the multiplier.

In an AP, the difference between two terms equals the difference of their indices multiplied by the common difference. Thus, \(a_{3n}-a_n=(3n-n)d=2nd\). Substituting the given values gives \(81-21=2n\times5\), so \(60=10n\). Therefore, \(n=6\). If \(n=5\), the difference would be \(2\times5\times5=50\), not 60. Exam tip: subtract the two term equations first; the first term \(a\) cancels automatically.

Use the arithmetic-progression relation a_n=a_k+(n−k)d. First evaluate the twelfth term from the given condition: a_12=5a_4−8=5(17)−8=85−8=77. Therefore a_12−a_4=77−17=60. The index gap from the fourth term to the twelfth term is 8, so 8d=60 and d=7.5. From a_4 to a_20 there are 20−4=16 equal steps. Hence a_20=a_4+16d=17+16(7.5)=17+120=137. Thus option A is correct. The other values would require a different common difference and would not preserve the relation between the fourth and twelfth terms. A fractional common difference is acceptable in an AP.