Finding the $n$th term of an AP

Each term in the sequence is 4 greater than the preceding term, so it is an arithmetic progression with first term 14 and common difference 4. The fourth term is \(a_4=a_1+3d=14+3\times4=26\). Remember that the common difference is added three times to reach the fourth term.

This is an arithmetic progression with common difference \(d=7-2=5\). Using \(a_5=a_1+(5-1)d\), we get \(a_5=2+4\times5=22\). Therefore, option C is correct. Exam tip: when the consecutive difference is constant, add the common difference to obtain the next term.

The terms are counted from left to right: the first term is 6, the second is 12, the third is 18, and the fourth is 24. Therefore, 24 is correct, which is option C. This is an arithmetic progression with common difference 6; in an exam, count the term position carefully.

Count the terms from left to right: the first term is 9, the second is 7, and the third is 5. Each successive term decreases by 2, so the third term is 5. Exam tip: Number the terms starting from 1 when identifying a term’s position.

This is an arithmetic progression with first term (12) and common difference (2). The fourth term is \(a_4=a_1+(4-1)d=12+3\times2=18\) cm, so option C is correct. For exams, remember that finding the fourth term requires adding the common difference three times to the first term, not four times.

The formula for the nth term of an AP is \(a_n=a+(n-1)d\). Thus, \(a_{10}=4+(10-1)\times3=4+27=31\). Therefore, 31 is correct. Choosing 30 results from an incorrect use of \(n-1\). Exam tip: for the 10th term, the common difference is added 9 times to the first term.

In this AP, the first term is \(a=7\) and the common difference is \(d=11-7=4\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{12}=7+(12-1)\times4=7+44=51\). Hence, 51 is correct. The number 55 is not the 12th term; it would be the 13th term. Exam tip: always use \(n-1\) in the formula for the \(n\)th term.

In this AP, the first term is 20 and the common difference is \(d=17-20=-3\). Using \(a_n=a+(n-1)d\), \(a_9=20+(9-1)(-3)=20-24=-4\). The option -1 can result from an incorrect subtraction. Exam tip: always retain the negative sign of the common difference in a decreasing AP.

The formula for the nth term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{15}=2+(15-1)\times5=2+70=72\). Hence, 72 is correct. The option 75 results from incorrectly adding \(15\times5\); only 14 common differences are added to reach the 15th term. Exam tip: always use \((n-1)\), not \(n\), in the nth-term formula.

Here (a=3), (d=5), so (a_{20}=3+19\times5=98). Keep the first term and term number correct in the formula.

The nth term of an arithmetic progression is given by \(a_n=a+(n-1)d\). Thus, \(a_{18}=9+(18-1)\times2=9+34=43\), so the correct answer is 43. The distractor 45 results from incorrectly using \(18d\) instead of \((18-1)d\). Exam tip: always use \(n-1\) when finding the nth term from the first term.

In this AP, the first term is \(a=50\) and the common difference is \(d=45-50=-5\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(a_7=50+(7-1)(-5)=50-30=20\). Therefore, 20 is correct. The number 25 is the sixth term, so it is a close but incorrect option. Exam tip: always retain the negative sign of \(d\) in a decreasing AP.

The formula for the nth term of an AP is \(a_n=a_1+(n-1)d\). Therefore, \(a_{11}=6+(11-1)\times7=6+70=76\). Hence, option D is correct. Exam tip: multiply the common difference by \(n-1\), not by \(n\), before adding the first term.

Here (a=-5), (d=3), so (a_{14}=-5+13\times3=34). Keep signs carefully with a negative first term.

Here (a=12), (d=6), so (a_{16}=12+15\times6=102). Start calculation from the given first term.

The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{13}=15+(13-1)(-2)=15-24=-9\). The option \(-7\) can result from using the term number or common difference incorrectly. Exam tip: for the \(n\)th term, always use \(n-1\) common differences, not \(n\).

In this arithmetic progression, the first term is \(a=1\) and the common difference is \(d=6-1=5\). Using \(a_n=a+(n-1)d\), we get \(a_{25}=1+(25-1)\times5=1+120=121\). The distractor 126 results from the common mistake of multiplying the common difference by 25 instead of 24. Exam tip: always use \(n-1\) in the formula for the nth term.

Here (a=10), (d=3), so (a_{30}=10+29\times3=97). The (30)th term contains (29) differences.

For this AP, the first term is \(a=100\) and the common difference is \(d=90-100=-10\). The nth-term formula is \(a_n=a+(n-1)d\). Hence, \(a_6=100+(6-1)(-10)=100-50=50\). Therefore, \(50\) is correct. \(40\) would be the 7th term. Exam tip: use \(n-1\), not \(n\), in the nth-term formula.

For an arithmetic progression, the formula is \(a_n=a_1+(n-1)d\). Here, \(a_1=25\), \(d=4\), and \(n=21\). Thus, \(a_{21}=25+(21-1)\times4=25+80=105\). Therefore, the correct answer is 105. Remember to multiply the common difference by \(n-1\), not by \(n\); this is why 109 is incorrect.