Finding the $n$th term of an AP

The difference between the 15th and 6th terms is \(9d\). Thus, \(9d=(x+65)-(x+11)=54\), so \(d=6\). From the 15th term to the 27th term, there are 12 common differences. Hence, \(a_{27}=x+65+12\times6=x+137\). Therefore, option B is correct. \(x+131\) would result from incorrectly using 11 common differences. Exam tip: Count the difference in term numbers carefully when moving between two AP terms.

\(a_{12}=74\) and \(8d=58\), so \(d=\frac{29}{4}\). \(a_{24}=16+20\cdot\frac{29}{4}=161\).

Using the two given terms, the common difference is \(d=\frac{128-38}{26-11}=\frac{90}{15}=6\). Now use \(a_n=a_{11}+(n-11)d\): \(218=38+(n-11)\times6\). Thus, \(180=6(n-11)\), so \(n-11=30\) and \(n=41\). Hence, 218 is the 41st term. The 40th term is \(212\), so it is a close but incorrect option. Exam tip: When two terms are given, first find \(d\) using the difference in their term numbers.

Given \(a_n=kn-8\), we have \(a_{19}=19k-8\) and \(a_7=7k-8\). Thus, \(a_{19}-a_7=(19k-8)-(7k-8)=12k=96\), so \(k=8\). Now \(a_{24}=24k-8=24\times8-8=184\). Hence, 184 is correct. Exam tip: when subtracting two terms, the constant term \(-8\) cancels out.

Here, the first term is 11 and the common difference is 8. Thus, \(a_n=11+8(n-1)=8n+3\). From \(8n+3>300\), we get \(n>37.125\), so the smallest possible integer is \(n=38\). Hence, \(a_{38}=307\) is the first term greater than 300. Although 299 is the nearest preceding term, it is not greater than 300. Exam tip: for “greater than” conditions, take the next valid integer value of \(n\).

Since p=8, a_p=a_8=29 and a_{p+10}=a_{18}=99. The difference over 10 terms is 99-29=70, so the common difference is d=70/10=7. Therefore, a_{35}=a_8+(35-8)d=29+27×7=218. Choosing 225 would incorrectly require a difference of 28 terms instead of 27 from the 8th term. Exam tip: when finding d from two terms, always use the difference between their indices.

In an AP, the sum of two terms equally spaced from a middle term equals twice that middle term. Since \(\frac{4+14}{2}=9\), we have \(a_4+a_{14}=2a_9\). Thus \(2a_9=98\), so \(a_9=49\). Choosing \(51\) would make the sum \(102\), not \(98\). Exam tip: Find the average of the term indices to identify the middle term quickly.

In an AP, terms with equally spaced indices have the middle term as their average. Hence, \(a_3+a_{13}=2a_8\). Therefore, \(a_3+a_8+a_{13}=3a_8=156\), giving \(a_8=52\). If \(a_8=54\), the sum would be 162, so it cannot be correct. Exam tip: For \(a_{n-r}, a_n, a_{n+r}\) in an AP, use \(a_{n-r}+a_{n+r}=2a_n\).

In an AP, \(a_{11}=a_1+10d\). Given \(a_{11}=a_1+80\), we get \(10d=80\), so \(d=8\). Therefore, \(a_{31}=a_1+30d=19+30\times8=259\). The value 251 would result from using an incorrect term position. Exam tip: in \(a_n=a_1+(n-1)d\), use \(n-1\), not \(n\).

The first term is \(a=7\), and the common difference is \(d=\frac{19}{2}-7=\frac{5}{2}\). Hence, \(a_{34}=a+(34-1)d=7+33\times\frac{5}{2}=\frac{179}{2}\). Therefore, \(\frac{179}{2}\) is correct. Getting 89 is a common error caused by not handling the denominator 2 correctly while adding 7. Exam tip: always use \(a_n=a+(n-1)d\), not \(a+nd\).

For an AP, \(a_n=a_1+(n-1)d\). Thus, \(a_{28}-a_{12}=16d=-96\), so \(d=-6\). Now \(a_{12}=a_1+11d=0\) gives \(a_1=-11(-6)=66\). Therefore, 66 is the correct option. If 60 were chosen, \(a_{12}\) would be \(-6\), not the given value. Exam tip: when two terms are given, first find \(d\) by subtracting the terms.

Given \(a_n=3n+4\), the common difference of the AP is \(d=3\). Moving 12 terms ahead from \(a_m=91\) increases the value by \(12\times3=36\). Therefore, \(a_{m+12}=91+36=127\). Option 124 would imply an increase of only 33, which is not correct for 12 steps. Exam tip: when the index increases by \(k\), an AP term increases by \(kd\).

The first term is 112 and the common difference is \(d=101-112=-11\). The relevant consecutive terms are \(35, 24, 13\). Since 24 is greater than 15 and the next term, 13, is less than 15, 13 is the first term below 15. Although 2 and -9 are also less than 15, they occur later. Exam tip: In a decreasing AP, check the two consecutive terms around the given boundary.

In an AP, the difference between two terms equals the difference of their positions multiplied by the common difference. Thus, \(a_{18}-a_6=(18-6)d=12d\). Given \(a_{18}-a_6=84\), we get \(12d=84\), so \(d=7\). The option 6 may result from counting the position gap incorrectly. Exam tip: use \(a_m-a_n=(m-n)d\) and subtract the indices carefully.

First find the common difference using the two given terms: \(d=\frac{a_{16}-a_4}{16-4}=\frac{78-18}{12}=5\). Now \(a_{4r}=a_4+(4r-4)d\). Thus, \(138=18+(4r-4)\times5\), giving \(4r-4=24\) and hence \(r=7\). If \(r=8\), the term would be \(a_{32}\), whose value is 158, not 138. Exam tip: For two non-consecutive AP terms, use \(d=\frac{a_n-a_m}{n-m}\).

The subscripts increase by 4 from 3 to 7 and from 7 to 11. Therefore, the corresponding term differences must be equal: (a_7-a_3)=(7x-12)-(3x-4)=4x-8. Hence, (a_{15}-a_{11}) is also 4x-8. So, (a_{15})=(11x-20)+(4x-8)=15x-28. The option (15x-26) has an incorrect constant term. Exam tip: In an AP, equal jumps in term numbers give equal differences between the terms.

Here, the first term is 20 and the common difference is 11. Thus, \(a_n=20+11(n-1)=11n+9\). For a term below 600, \(11n+9<600\), so \(n<53.73\). Hence the greatest possible integer value is \(n=53\), giving \(a_{53}=11\times53+9=592\). The next term, 603, is greater than 600. Exam tip: In range questions, use an inequality first and then take the greatest valid integer value of n.

Using \(a_p=a+(p-1)d\) and \(a_q=a+(q-1)d\) gives \((p-q)d=0\). Since \(p\ne q\), \(d=0\), so every term is equal. Exam tip: isolate the difference of indices first.

Let the first term be \(a\) and the common difference be \(d\). Then \(a_2+a_6=(a+d)+(a+5d)=2a+6d=50\), so \(a+3d=25\). Also, \(a_{10}=a+9d=57\). Subtracting the two equations gives \(6d=32\), hence \(d=\frac{16}{3}\). Now \(a_{22}=(a+3d)+18d=25+18\times\frac{16}{3}=121\). Therefore, 121 is correct. A value such as 105 can result from using an incorrect common difference. Exam tip: subtract equations involving AP terms first to find \(d\) quickly.

For \(a_n=pn+q\), \(a_{n+1}-a_n=[p(n+1)+q]-(pn+q)=p\), a constant. Hence it is an AP. In \(pn^2+q\), the difference changes with n. Exam tip: a linear nth-term form indicates an AP.