Finding the $n$th term of an AP

For two terms of an AP, \(a_{18}=a_4+(18-4)d\). Therefore, \(a_4=130-14\times 8=130-112=18\). Hence, 18 is correct. Option 20 is close, but adding 14 common differences to it gives \(a_{18}=132\), not 130. Exam tip: to find an earlier term from a later term, subtract the position gap multiplied by \(d\).

The station distances form an AP with first term \(a=12\) km and common difference \(d=7\) km. The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{15}=12+(15-1)\times7=12+98=110\) km. Hence, 110 km is correct. The value 117 km would result from incorrectly adding \(15\times7\); there are only 14 gaps after the first station. Exam tip: always use \((n-1)\) in the nth-term formula.

The scores form an arithmetic progression with first term \(a=20\), common difference \(d=6\), and \(n=18\). The nth term is \(a_n=a+(n-1)d\). Therefore, \(a_{18}=20+(18-1)\times6=20+102=122\). Hence, 122 points are earned at the 18th level. Getting 120 usually results from using the wrong number of intervals. Exam tip: for the nth term, add \(n-1\) common differences, not \(n\).

This is an arithmetic progression with first term \(a=32\), common difference \(d=4\), and \(n=25\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{25}=32+(25-1)\times4=32+96=128\). Hence, 128 is correct. Adding 4 twenty-five times would give 132, but the increase occurs only 24 times after the first shelf. Exam tip: always use \((n-1)\), not \(n\), in the nth-term formula.

The water amounts form a decreasing AP, with first term \(a=240\) and common difference \(d=-18\). Thus, \(a_9=a+(9-1)d=240+8(-18)=96\) litres. The value \(102\) litres would count only 7 decreases, whereas there are 8 decreases from the first hour to the ninth hour. Exam tip: In \(a_n=a+(n-1)d\), use \(n-1\), not \(n\).

This forms an arithmetic progression with first term \(a=40\), common difference \(d=3\), and \(n=30\). Using \(a_n=a+(n-1)d\), we get \(a_{30}=40+(30-1)\times3=40+87=127\). Therefore, 127 push-ups will be done on the 30th day. The option 130 results from incorrectly using \(n\) instead of \(n-1\). Exam tip: From the first term to the \(n\)th term, the common difference is added \(n-1\) times.

This is an AP with first term \(a=1500\), common difference \(d=75\), and \(n=14\). Using \(a_n=a+(n-1)d\), \(a_{14}=1500+(14-1)\times75=1500+975=₹2475\). Option ₹2550 incorrectly adds the increase 14 times; by the 14th month, it has been added only 13 times. Exam tip: use \((n-1)d\), not \(nd\), for the \(n\)th term.

The position gap is (18-11=7) and (d=13) so the difference is (7\times13=91). You do not need to find both terms separately.

For an AP, \(a_n=a+(n-1)d\). Thus, \(a_{22}-a_{10}=12d=130-58=72\), so \(d=6\). Now \(a_{10}=a+9d\) gives \(58=a+9\times6\), hence \(a=4\). Option 6 is the common difference, not the first term. Exam tip: When two terms are given, first use the difference in their term numbers to find \(d\).

Here, the first term is \(a=z\) and the common difference is \(d=5\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(112=z+(19-1)\times5=z+90\), so \(z=22\). If 20 were used, the 19th term would be \(20+90=110\), not 112. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

From (-12=73+(n-1)(-5)), (85=5(n-1)) so (n=18). In a decreasing AP keep signs correct up to the negative target.

(d=\frac{103-47}{15-8}=8) so (a_{22}=103+7\times8=159). Add equal term gaps for equally spaced positions.

The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{15}=12+(15-1)\times\frac{7}{2}=12+14\times\frac{7}{2}=12+49=61\). Hence, 61 is correct. A value such as 59 can result from using an incorrect number of common differences. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

Given \(a_n=10-3n\), set the required term equal to \(-44\): \(-44=10-3n\). Thus, \(-54=-3n\), so \(n=18\). Therefore, \(-44\) is the 18th term of the AP. The 17th term is \(a_{17}=10-51=-41\), so it is not correct. Exam tip: To find a term number, equate the given value to \(a_n\) and solve for \(n\).

(d=\frac{113-65}{8}=6) and (a_6=65-7\times6=23). When moving backward from a known term subtract (d).

From (189=18+(n-1)9), (171=9(n-1)) so (n=20). For term number subtract the first term and divide by (d).

The ((r+6))th term is (6d) ahead of the (r)th term so (3r-2+24=3r+22). In symbolic questions look at the position gap.

From the 5th term to the 14th term, there are 9 equal gaps. Therefore, the common difference is \(d=\frac{90-27}{14-5}=\frac{63}{9}=7\). The 20th term is 6 places after the 14th term, so \(a_{20}=90+6\times7=132\). The value 126 would result from moving only 5 places and would correspond to the 19th term. Exam tip: when moving between two AP terms, count the gaps as \(n-m\).

Here (a=408) and (d=12) so (a_{18}=408+17\times12=612). Choose the first correct multiple after the limit.

In (13n<600), the greatest (n=46) so the term is (13\times46=598). Take the greatest integer below the limit.