Finding the $n$th term of an AP

Here (a=-28) and (d=8) so (a_{21}=-28+20\times8=132). Be careful while adding a negative first term.

In an AP, the difference between two terms equals the difference in their positions multiplied by the common difference. Thus, \(a_{16}-a_7=(16-7)d\). Hence, \(90-36=9d\), so \(d=6\). If the difference were 7, the difference across 9 positions would be \(63\), not the given \(54\). Exam tip: When two terms of an AP are known, use \(d=\frac{a_n-a_m}{n-m}\) directly.

(d=\frac{101-38}{15-6}=7) so (a_{24}=101+9\times7=164). Moving forward from the nearer known term is easier.

Here, the first term is \(a=120\) and the common difference is \(d=109-120=-11\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Thus, \(a_{13}=120+(13-1)(-11)=120-132=-12\). Therefore, \(-12\) is correct. \(-10\) can result from incorrectly counting the number of differences or using \(n\) instead of \((n-1)\). Exam tip: for the \(n\)th term, always use \((n-1)\) common differences.

For an AP, \(a_n=a+(n-1)d\). Substituting \(a=31\), \(n=19\), and \(a_{19}=139\), we get \(139=31+18d\). Thus, \(108=18d\), so \(d=6\). If the common difference were 5, the 19th term would be \(31+18\times5=121\), not 139. Exam tip: the common difference is added \(n-1\) times to reach the \(n\)th term.

To find a term from a known term of an AP, use \(a_n=a_m+(n-m)d\). Thus, \(a_{19}=a_5+(19-5)d=24+14\times8=24+112=136\). Hence, 136 is correct. Getting 144 would indicate that the difference between the term positions was taken incorrectly. Exam tip: the number of common differences between \(a_m\) and \(a_n\) is always \(n-m\).

Here (a=2.4) and (d=1.6) so (a_{16}=2.4+15\times1.6=26.4). Keep place value in mind while multiplying decimals.

The sequence is given directly by the nth-term rule a_n=6n+7. To find the 34th term, substitute n=34: a_34=6(34)+7=204+7=211. Therefore option B is correct. The nearby alternatives result from arithmetic or substitution errors; for example, 6(34)=204, not 200 or 208, and the final addition must include the constant 7.

Given \(a_n=95-5n\), substitute \(n=26\): \(a_{26}=95-5\times26=95-130=-35\). Hence, option A is correct. \(-30\) would result from an error in multiplication or subtraction. Exam tip: after substituting \(n\), multiply first and then subtract.

Here (a=\frac{2}{3}) and (d=\frac{2}{3}) so (a_{28}=\frac{2}{3}+27\cdot\frac{2}{3}=\frac{56}{3}). Simplify multiplication first in fractions.

For an AP, \(a_n=a_1+(n-1)d\). Thus, \(92=a_1+(14-1)\times7=a_1+91\), so \(a_1=1\). Option \(85\) would result from subtracting \(d\) only once, but moving from the 14th term to the first term requires subtracting \(13d\). Exam tip: always use \(n-1\) in the formula for the \(n\)th term.

From (206=26+(n-1)9), (180=9(n-1)) so (n=21). Divide the difference between the term and first term by (d).

(a=-45) and (d=9) so (a_{20}=-45+19\times9=126). In an increasing AP starting negative add carefully at the end.

(d=\frac{108-52}{17-9}=7) so (a_{25}=108+8\times7=164). Equal position gaps have equal term gaps in an AP.

For an AP, \(a_n=a+(n-1)d\). Here, \(a=22\), \(n=32\), and \(a_{32}=177\). Thus, \(177=22+31d\), so \(31d=155\) and \(d=5\). If the common difference were 4, the 32nd term would be \(22+31\times4=146\), not 177. Exam tip: the common difference is added \(n-1\) times to obtain the \(n\)th term.

Here, the first term is \(a=105\) and the common difference is \(d=98-105=-7\). Therefore, \(a_n=a+(n-1)d=105-7(n-1)=112-7n\). For a negative term, \(112-7n<0\), so \(n>16\). The 16th term is \(0\), which is not negative; hence the 17th term, \(-7\), is the first negative term. Exam tip: For a negative term use \(<0\), not \(\leq0\).

The terms are of the form (13+8(n-1)) and the greatest such term less than (250) is (245). In limit questions check the next term too.

Putting (n=1), (a_1=15) and the coefficient of (n) is (d=4). The direct formula gives both values quickly.

From (-1=63+(n-1)(-4)), (64=4(n-1)) so (n=17). Handle signs carefully with a negative target term.

Here, the first term is 7 and the common difference is 5. The nth term is \(a_n=7+5(n-1)\). From \(7+5(n-1)<180\), we get \(n<35.6\), so the greatest integer value of n is 35. Thus, \(a_{35}=7+5(34)=177\). The next term is 182, which is not less than 180. Although 179 is less than 180, it is not a term of this AP. Exam tip: For a “last term less than” question, solve the inequality and take the greatest possible integer value of n.