Finding the $n$th term of an AP

From (-17=55+(n-1)(-4)), (72=4(n-1)) so (n=19). Keep signs correct while reaching a negative term.

(d=\frac{77-42}{7}=5) so (a_{23}=77+7\times5=112). Equally spaced terms have equal increases.

The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{17}=9+(17-1)\times\frac{5}{2}=9+16\times\frac{5}{2}=9+40=49\). Hence, the correct answer is 49. Getting 47 would result from using the number of differences incorrectly. Exam tip: for the \(n\)th term, use \(n-1\) common differences, not \(n\).

Given \(a_n=7-2n\), set the nth term equal to \(-35\): \(-35=7-2n\). Thus, \(-42=-2n\), so \(n=21\). Hence, \(-35\) is the 21st term of the AP. The 20th term is \(7-2(20)=-33\), so option C is not correct. Exam tip: To find the term number for a given value, equate that value to \(a_n\) and solve for \(n\).

(d=\frac{72-40}{8}=4) and (a_4=40-7\times4=12). When moving backward from a known term, subtract (d).

Here, the first term is \(a=17\) and the common difference is \(d=24-17=7\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Thus, \(136=17+(n-1)\times7\), so \(119=7(n-1)\). Hence, \(n-1=17\) and \(n=18\). Option 17 is incorrect because the 17th term is \(129\). Exam tip: always use \((n-1)\), not \(n\), in the AP nth-term formula.

The ((m+5))th term is (5d) ahead of the (m)th term, so (2m+3+10=2m+13). In symbolic terms, look at the position gap.

Given \(a_4=15\) and \(a_{12}=55\). Thus, \(a_{12}-a_4=8d\), so \(55-15=8d\) and \(d=5\). Now \(a_{16}=a_{12}+4d=55+4\times5=75\). Therefore, 75 is correct. The nearby option 80 would result from adding \(5d\) instead of the required \(4d\). Exam tip: use the difference between the term numbers to determine how many common differences are involved.

Here (a=304) and (d=8) so (a_{25}=304+24\times8=496). Choose the first correct multiple after the limit.

In (11n<500), the greatest (n=45), so the term is (11\times45=495). Take the greatest integer below the limit.

The given formula is \(a_n=7n+12\). For the \(41\)st term, substitute \(n=41\): \(a_{41}=7\times41+12=287+12=299\). Therefore, the correct answer is \(299\). A choice such as \(294\) may result from an arithmetic error. Exam tip: substitute the given value of \(n\) directly into the formula for \(a_n\).

Here (d=9) so (a_{20}=48+14\times9=174). (a_{n+1}-a_n) is the common difference of the AP.

In an AP, \(a_1=a\) and \(a_2=a+d\). Hence, \(a_1+a_2=a+(a+5)=33\), so \(2a=28\) and \(a=14\). Therefore, \(a_{12}=a+11d=14+11\times5=69\). Choosing 71 would result from using the number of terms or the common difference incorrectly. Exam tip: while finding the \(n\)th term, use \(a_n=a+(n-1)d\), ensuring that the multiplier is \(n-1\).

(a_{10}) is the middle term between (a_6) and (a_{14}), so (a_{10}=\frac{100}{2}=50). The average of equally spaced terms is the middle term.

Here, the first term is \(a=90\) and the common difference is \(d=-7\). The terms proceed as \(90,83,76,\ldots,27,20\). The next term after \(27\) is \(20\), which is not greater than \(20\). Therefore, the last term greater than \(20\) is \(27\). Exam tip: In a decreasing AP, check the two consecutive terms around the given limit.

Here (a=6) and (d=8) so (a_{23}=6+22\times8=182). Remember to use (n-1) instead of (n) in exams.

Here (d=-6) so (a_{18}=84+17(-6)=-18). In a decreasing AP take the common difference as negative.

From (170=17+(n-1)9), (153=9(n-1)) so (n=18). Add (1) at the end while finding the term number.

For an AP, \(a_n=a+(n-1)d\). Thus, \(71=a+11\times5\), so \(a=71-55=16\). If 18 were taken as the first term, the 12th term would be \(18+11\times5=73\), not 71. Exam tip: use \((n-1)\), not \(n\), in the formula for the nth term.

From (201=9+(n-1)8), (192=8(n-1)) and (n=25). If the term number is an integer the answer is on the right track.