Finding the $n$th term of an AP

(a_{24}) is equally spaced between (a_{12}) and (a_{36}). Therefore (a_{24}=\frac{75+219}{2}=147).

The difference between consecutive terms is \(7\), so the first term is \(a=t-8\) and the common difference is \(d=7\). The \(25\)th term is \(a_{25}=a+24d\). Thus, \(230=(t-8)+24\times7=t+160\), giving \(t=70\). If \(t=66\), the \(25\)th term would be \(226\), so it is not correct. Exam tip: in the formula for the \(n\)th term, multiply \(d\) by \(n-1\), not by \(n\).

Given \(a_n=11n-5\), we have \(a_{4k}=11(4k)-5=44k-5\) and \(a_k=11k-5\). Therefore, \(a_{4k}-a_k=(44k-5)-(11k-5)=33k\). From \(33k=198\), we get \(k=6\). For example, if \(k=5\), the difference would be only 165, not 198. Exam tip: Substitute each index separately in the nth-term formula before subtracting.

The nth term of an AP is \(a_n=a_1+(n-1)d\). Substituting \(a_1=11\) and \(d=-4\) gives option A. In option B, putting \(n=1\) gives 7, not 11. Exam tip: always test \(n=1\) to verify the first term.

From (145=119+c), (c=26). (757=51r+26), giving (r=\frac{731}{51}), so option checking is necessary.

The sequence of positive multiples is \(23, 46, 69, \ldots\), and its \(n\)th term is \(23n\). For the last term, \(23n<2000\). Since \(2000\div23\approx86.95\), the greatest integer value of \(n\) is 86. Thus, the last term is \(23\times86=1978\). Although \(2001=23\times87\), it is greater than 2000. Exam tip: for “less than” questions, check that the next multiple does not cross the given limit.

The first term is \(a=703\) and the common difference is \(d=19\), since each successive multiple increases by 19. The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(a_{24}=703+(24-1)\times19=703+437=1140\). Therefore, 1140 is correct. The value 1159 is the next, or 25th, term. Exam tip: do not forget to use \(n-1\) in the nth-term formula.

For an AP, \(a_n=a+(n-1)d\). Thus, \(a_6+a_{18}=2a+22d=156\) and \(a_{10}+a_{22}=2a+30d=252\). Subtracting the first equation from the second gives \(8d=96\), so \(d=12\). Now \(2a+22(12)=156\) gives \(a=-54\). Hence, \(a_{34}=a+33d=-54+33(12)=342\). The value \(324\) does not satisfy the two given conditions. Exam tip: subtract two such sum equations to eliminate \(a\) and find \(d\) quickly.

In an AP, the difference of consecutive terms is the common difference \(d\), so \(a_{n+1}-a_n=d\) must be constant. A constant ratio instead identifies a GP. Exam tip: subtract two consecutive nth-term expressions to test for an AP.

Here, the first term is \(a=35\) and the common difference is \(d=11\). The \(n\)th term is \(a_n=35+11(n-1)\). Using \(35+11(n-1)<600\), we get \(n<52.36\ldots\), so the greatest possible integer value is \(n=52\). Check: \(a_{52}=596<600\), whereas \(a_{53}=607>600\). Therefore, 52 terms are less than 600. Exam tip: For a ‘less than’ condition, verify the next term to confirm the count.

Putting \(n=1\) gives \(a_1=7-3=4\). In \(a_n=a+(n-1)d\), the coefficient of \(n\) is the common difference, so \(d=-3\). Exam tip: always verify the first term by substituting \(n=1\).

In an AP, the common difference is the same between consecutive terms. Here, \(a_9=a_4+5d\) and \(a_{14}=a_4+10d\). Therefore, \(20+5d+20+10d=160\), so \(15d=120\) and \(d=8\). Now, \(a_{28}=a_4+24d=20+24\times8=212\). Hence, 212 is correct. The value 204 does not result when the correct common difference \(d=8\) is used. Exam tip: when one term \(a_r\) is known, use \(a_n=a_r+(n-r)d\) directly.

Here, the first term is \(a=-38\) and the common difference is \(d=13\). Thus, \(a_n=-38+13(n-1)=13n-51\). For \(a_n>300\), we get \(13n-51>300\), so \(n>27\). The least integer value is \(n=28\), and \(a_{28}=313\). Note that 300 is not itself a term of the AP, so 313 is the correct answer. Exam tip: for “greater than,” use \(>\) and then take the least possible integer value of \(n\).

Given \(a_n=6n+q\), we get \(a_{5n}=6(5n)+q=30n+q\). Hence, \(a_{5n}-a_n=(30n+q)-(6n+q)=24n\). Now \(24n=216\), so \(n=216/24=9\). Therefore, the correct option is \(9\). The term \(q\) cancels because it occurs in both terms. Exam tip: In such questions, first write \(a_{5n}\) explicitly and then subtract \(a_n\).

The indices 7, 10 and 13 are equally spaced by 3, so their AP terms are also equally spaced. Therefore, the middle term is the average: \(a_{10}=\frac{a_7+a_{13}}{2}\). Exam tip: look for equally spaced term numbers.

(a_{2p}-a_p=pd=48), so (6p=48) and (p=8). In index-based questions, first find the difference of positions.

(d=\frac{95-23}{18-6}=6), so (a_{33}=95+15\times6=185). First find (d), then move from the nearer known term.

(12k=96), so (k=8) and (a_{29}=8\times29-7=225). In a direct formula, find the coefficient first.

Here (d=-7). After (-94), the next term is (-101), so it is the first term less than (-100).

There are 12 term-intervals between the 8th and the 20th terms. Hence, \(12d=(x+91)-(x+19)=72\), so \(d=6\). From the 20th term to the 35th term, there are 15 intervals; therefore, \(a_{35}=x+91+15\times6=x+181\). Thus, \(x+181\) is correct. \(x+175\) would result from incorrectly using only 14 intervals. Exam tip: first find \(d\) from the two given terms, then count the term-intervals from the nearer known term.