Finding the $n$th term of an AP

The first row is term 1, so the increase of 3 occurs only 11 times before term 12: \(18+11\times3=51\). Using \(18+12\times3\) counts one extra increase. In exams, always use \(n-1\).

The first term of this AP is 4 and the common difference is 0, since \(4-4=0\) for consecutive terms. Using \(a_n=a+(n-1)d\), \(a_{50}=4+49\times0=4\). Choosing 50 confuses the term number with the value of the term. Exam tip: find the common difference first; if \(d=0\), every term equals the first term.

In this AP, the first term is \(a=6\) and the common difference is \(d=10-6=4\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{31}=6+(31-1)\times4=6+120=126\). Choosing 122 would correspond to adding only 29 differences, so it is close to the 30th term. Exam tip: use \(n-1\), not \(n\), when finding the \(n\)th term.

In this AP, the first term is \(a=9\) and the common difference is \(d=5-9=-4\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_8=9+(8-1)(-4)=9-28=-19\). Hence, option B is correct. \(-15\) is the seventh term, since \(a_7=9+6(-4)=-15\). Exam tip: use \(n-1\) in the nth-term formula and put a negative common difference in brackets.

The nth term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{13}=3+(13-1)\times10=3+120=123\). Hence, 123 is correct. The value 120 is only \(12d\); the first term, 3, must also be added. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

The relevant rule is aₙ = a + (n − 1)d for the nth term of an AP. The first term is a = 11, and the common difference is d = 16 − 11 = 5. For n = 9, a₉ = 11 + (9 − 1)×5 = 11 + 8×5 = 11 + 40 = 51. Therefore option D is correct. Only eight differences are added because the first term is already term one; moving from term one to term nine requires eight steps. Option A stops too early, option B uses an incorrect number of increments, and option C adds two extra steps. The sequence 11, 16, 21, ..., 51 also confirms the calculation.

Given \(a_n=2n+3\), substitute \(n=18\) to find the 18th term: \(a_{18}=2(18)+3=36+3=39\). Therefore, 39 is correct. The option 41 can result from an arithmetic error while multiplying or adding. Exam tip: substitute the term number carefully into the formula before simplifying.

The general term is \(a_n=5n-1\). To find the 20th term, substitute \(n=20\): \(a_{20}=5(20)-1=100-1=99\). Therefore, 99 is correct. The value 101 would result from adding 1 instead of subtracting it. Exam tip: In an \(n\)th-term question, first substitute the required term number for \(n\).

Given \(a_n=7n+2\), substitute \(n=11\) to get \(a_{11}=7\times11+2=77+2=79\). Therefore, 79 is correct. The close distractor 77 results from forgetting to add \(+2\). In exams, substitute the term number carefully for \(n\).

To find the sixth term, substitute \(n=6\) in the given general term: \(a_6=30-3(6)=30-18=12\). Hence, the correct answer is 12. The value 18 is only \(3\times6\); it must still be subtracted from 30. Exam tip: Always substitute the term number carefully for \(n\) in \(a_n\).

The nth term of an AP is \(a_n=a_1+(n-1)d\). Therefore, \(a_{10}=13+(10-1)\times 8=13+72=85\). Hence, 85 is correct. The value 93 results from incorrectly adding \(10d\) instead of \((10-1)d\). Exam tip: for the nth term, always use \((n-1)\) common differences.

For an arithmetic progression, the nth term is given by aₙ = a + (n − 1)d. Here a = 18 and d = 22 − 18 = 4. Substituting n = 24 gives a₂₄ = 18 + (24 − 1)×4 = 18 + 23×4 = 18 + 92 = 110. Therefore option B is correct. The multiplier is 23 because the first term is counted as the starting value and only 23 equal increases are needed to reach the 24th term. Option A results from using too few increments, option C from using one extra increment, and option D does not reflect the sequence's steady increase. The answer can also be checked by adding 4 successively from the first term.

The nth term of an AP is \(a_n=a+(n-1)d\). Thus, \(a_5=40+(5-1)(-6)=40-24=16\). Hence, the correct answer is 16. Getting 20 would mean subtracting the common difference only three times. Exam tip: always use \((n-1)\) common differences for the nth term.

Here (a=100), (d=50), so (a_8=100+7\times50=450). The question asks only the (8)th month's deposit, not the total.

In this AP, the first term is \(a=21\) and the common difference is \(d=28-21=7\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{14}=21+(14-1)\times7=21+91=112\). Hence, \(112\) is correct. \(119\) is the next, or 15th, term. Exam tip: for the \(n\)th term, add the common difference \(n-1\) times.

The governing concept is the nth-term formula for an arithmetic progression: a_n = a + (n − 1)d. The first term is a = 2.5, and the common difference is d = 5 − 2.5 = 2.5. To reach the ninth term, we need eight equal jumps after the first term. Therefore, a_9 = 2.5 + (9 − 1)(2.5) = 2.5 + 8 × 2.5 = 2.5 + 20 = 22.5. Hence option B is correct. Option A can result from mishandling the initial term, while option C adds one extra common difference. Option D does not follow the constant step of 2.5. The decimal form causes no difficulty because the nth-term formula works equally with whole numbers and decimals.

Here (a=\frac{1}{2}), (d=\frac{1}{2}), so (a_{16}=\frac{1}{2}+15\times\frac{1}{2}=8). In fractions, keep denominators clear.

The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{26}=0+(26-1)\times4=25\times4=100\). Hence, the correct answer is \(100\). \(104\) would result from incorrectly adding 26 common differences. Exam tip: always use \((n-1)\) in the formula for the \(n\)th term.

Here (a=-3), (d=-4), so (a_{12}=-3+11(-4)=-47). Check signs twice when both values are negative.

For this AP, the first term is \(a=45\) and the common difference is \(d=39-45=-6\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Thus, \(a_{10}=45+(10-1)(-6)=45-54=-9\). Therefore, \(-9\) is correct. Choosing \(-3\) results from using the number of differences incorrectly. Exam tip: for the \(n\)th term, use \(n-1\) differences, not \(n\).

The nth term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{40}=17+(40-1)\times1=17+39=56\). Hence, 56 is correct. Getting 55 is an off-by-one error, because 39 common differences are added after the first term. Exam tip: always use \(n-1\) when finding the nth term.

This is an nth-term problem for an arithmetic progression. The first term is a = 16, and the common difference is d = 24 − 16 = 8. The governing formula is a_n = a + (n − 1)d because the first term is followed by n − 1 equal jumps. Substituting n = 18 gives a_18 = 16 + (18 − 1) × 8 = 16 + 17 × 8 = 16 + 136 = 152. Therefore option D is correct. A frequent error is to use 18 × 8 without correctly accounting for the starting term and the number of jumps. The value 144 may arise from multiplying the term number by the difference, while 160 comes from adding an incorrect extra step. Exactly seventeen differences are required after the first term.

The general term is \(a_n=4n+6\). Substituting \(n=25\), we get \(a_{25}=4\times25+6=100+6=106\). Hence, 106 is correct. The value 100 is only \(4\times25\); it misses the constant term 6. Exam tip: while finding a term, substitute the given term number carefully for \(n\).

The nth term of an AP is \(a_n=a+(n-1)d\). Thus, \(a_{20}=5+(20-1)(-1)=5-19=-14\). Therefore, \(-14\) is correct. The value \(-13\) would result from incorrectly using 18 in place of \(n-1\). Exam tip: always multiply the common difference by \(n-1\) when finding the nth term.

Here (a=1), (d=2), so (a_{51}=1+50\times2=101). Even for odd numbers, the formula is the safest method.