Finding the $n$th term of an AP

The first term of the AP is \(a=-4\), and its common difference is \(d=1-(-4)=5\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{31}=-4+(31-1)\times5=-4+150=146\). Hence, 146 is the correct answer. Using \(31\times5\) instead of \((31-1)\times5\) would give 151, which is incorrect. Exam tip: always use \(n-1\) in the formula for the \(n\)th term.

The difference of terms is (81-41=40) and the difference of positions is (10), so (d=4). Use (d=\frac{\text{difference of terms}}{\text{difference of positions}}).

(d=\frac{46-14}{12-4}=4), so (a_{20}=46+8\times4=78). Moving from the nearest known term is a faster method.

In this AP, the first term is \(a=100\) and the common difference is \(d=92-100=-8\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{15}=100+(15-1)(-8)=100-112=-12\). Hence, \(-12\) is correct. An answer such as \(-4\) can result from using the wrong term count instead of \(n-1\). Exam tip: always check the negative sign of the common difference.

For an AP, the nth-term formula is \(a_n=a+(n-1)d\). Here, \(a=18\) and \(a_{16}=93\), so \(93=18+15d\). Thus, \(15d=75\), giving \(d=5\). If the common difference were 4, the 16th term would be \(18+15\times4=78\), not 93. Exam tip: use \(n-1\), not \(n\), in the nth-term formula.

For any two terms of an AP, \(a_n=a_m+(n-m)d\). Therefore, \(a_{17}=a_3+(17-3)d=11+14\times6=95\). Hence, 95 is correct. Option 83 uses \(12\times6\), but there are 14 common differences from the 3rd term to the 17th term. Exam tip: always use the difference of the term indices, \(n-m\).

Here (a=2.5), (d=2.5), so (a_{20}=2.5+19\times2.5=50). In decimal APs, keep multiplication exact.

Given \(a_n=4n-1\), substitute \(n=30\) to find the 30th term: \(a_{30}=4(30)-1=120-1=119\). Therefore, 119 is correct. The value 121 would result from adding 1 instead of subtracting it, so it is not correct. Exam tip: in a direct \(n\)th-term formula, substitute the required term number carefully for \(n\).

Using the given direct formula \(a_n=7-3n\), substitute \(n=22\): \(a_{22}=7-3(22)=7-66=-59\). Hence, \(-59\) is correct. An answer such as \(-57\) can result from an error in multiplying \(3\times22\) or in subtraction. Exam tip: In a direct formula, substitute the term number for \(n\) and perform multiplication before subtraction.

Here (a=\frac{1}{2}), (d=\frac{1}{2}), so (a_{40}=\frac{1}{2}+39\cdot\frac{1}{2}=20). Fractions can also be checked as decimals.

For an AP, \(a_n=a_1+(n-1)d\). Thus, \(35=a_1+(8-1)\times4=a_1+28\), so \(a_1=7\). If 9 were chosen, the eighth term would be \(9+28=37\), not 35. Exam tip: always use \((n-1)d\) in the formula for the \(n\)th term.

Here, the first term is \(a=15\) and the common difference is \(d=21-15=6\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(111=15+(n-1)\times6\), so \(96=6(n-1)\) and \(n=17\). Therefore, 111 is the 17th term. The 16th term is \(105\), so it is close but not correct. Exam tip: while finding a term number, remember that the formula contains \((n-1)\).

Here (a=-20), (d=6), so (a_{19}=-20+18\times6=88). In an increasing AP starting negative, add carefully at the end.

(d=\frac{32-16}{9-5}=4), so (a_{13}=32+4\times4=48). Equal position gaps give equal term gaps in an AP.

The nth-term formula of an AP is \(a_n=a+(n-1)d\). Here, \(a=9\) and \(a_{21}=69\), so \(69=9+(21-1)d=9+20d\). Hence, \(20d=60\), giving \(d=3\). If 2 were used, the 21st term would be \(9+20\times2=49\), not 69. Exam tip: the nth term contains \((n-1)d\), not \(nd\).

Here, the first term is \(a=45\) and the common difference is \(d=-5\). Therefore, \(a_n=a+(n-1)d=45-5(n-1)=50-5n\). For the first negative term, \(50-5n<0\), so \(n>10\). The smallest integer satisfying this is \(11\); hence, the 11th term is the first negative term. The 10th term is \(0\), which is not negative. Exam tip: For a “first” term question, choose the smallest positive integer that satisfies the inequality.

Here, the first term is \(a=6\) and the common difference is \(d=7\). Thus, the terms increase by 7: \(\ldots, 76, 83, 90\). Therefore, the last term just less than \(90\) is \(83\). The numbers \(84\), \(85\), and \(86\) are not terms of this AP. Exam tip: To identify the term immediately before a given number, also verify the next AP term.

Putting (n=1), (a_1=7), and consecutive terms differ by (2). In a direct formula, (d) is the coefficient of (n).

From (0=30+(n-1)(-2)), (2(n-1)=30), so (n=16). Use the same nth-term formula even for the zero term.

Here, the first term is 1 and the common difference is 3, so the nth term is \(a_n=1+3(n-1)=3n-2\). From \(3n-2<100\), we get \(n<34\); hence the greatest integral value of n is \(33\). Therefore, \(a_{33}=97\). Although 99 is less than 100, it is not a term of this AP because the terms increase by 3 from 1. Exam tip: write \(a_n<\) the given number and find the greatest possible integer value of n.