Finding the $n$th term of an AP

The given formula is \(a_n=8n+15\). Substituting \(n=36\), we get \(a_{36}=8\times36+15=288+15=303\). Therefore, 303 is the correct answer. Getting 301 would indicate an error in multiplication or addition. Exam tip: In an AP nth-term formula, directly substitute the required term number for \(n\).

Since \(a_{n+1}-a_n=11\), the common difference is \(d=11\). There are 10 term-steps from \(a_7\) to \(a_{17}\), so \(a_{17}=a_7+(17-7)d=58+10\times11=168\). Hence, option D is correct. A value such as 166 results from an error in counting the term difference or multiplying. Exam tip: use \(a_n=a_m+(n-m)d\) when one term of an AP is known.

Let the first term be \(a_1=a\). Then \(a_2=a+d=a+7\). Hence, \(a_1+a_2=a+(a+7)=41\), so \(2a=34\) and \(a=17\). Therefore, \(a_{10}=a+9d=17+9\times7=80\). Choosing 83 would result from using the term number incorrectly instead of \(n-1\). Exam tip: always use \(a_n=a+(n-1)d\) for the \(n\)th term of an AP.

(a_{13}) is the middle term between (a_7) and (a_{19}) so (a_{13}=\frac{156}{2}=78). The average of equally spaced terms is the middle term.

The first term of the AP is 128 and the common difference is \(d=-9\). The term after 38 is \(38-9=29\). Since \(38>30\) but \(29<30\), 38 is the last term greater than 30. Option 29 is a close distractor, but it is not greater than 30. Exam tip: In a decreasing AP, check the next term to confirm that it crosses the given limit.

(d=\frac{103-31}{19-7}=6), so (a_{34}=103+15\times6=193). First find (d), then move forward from the nearer term.

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 \(25-1\).

From (50=36+p), (p=14) and (a_{27}=81+14=95). First find the constant, then substitute the required term number.

(a_{2m}-a_m=md=27), so (3m=27) and (m=9). In an AP, term difference is linked to position difference.

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.

(a_2), (a_5), and (a_8) are equally spaced, so the middle term is the average. (a_5=\frac{111}{3}=37).

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.