Finding the $n$th term of an AP

(a_{16}=90) and (9d=68) so (d=\frac{68}{9}). (a_{34}=22+27\cdot\frac{68}{9}=226).

The \(n\)th term of an AP has the linear form \(an+b\), so consecutive differences must be constant. For \(n^2+1\), the differences are \(3,5,7,\ldots\), not constant. Exam tip: identify whether the formula is linear in \(n\).

From (128=99+c), (c=29). (392=44r+29) does not give an integer, so (a_{4r}=381) would give (r=8).

The terms are (33+15(n-1)). The first term greater than (1000) is (1008) because the previous term is (993).

In an AP, the difference between consecutive terms is constant. Here, \(a_{n+1}-a_n=[4(n+1)-9]-(4n-9)=4\), so it is an AP. A constant ratio identifies a GP instead. Exam tip: check the coefficient of \(n\) for the common difference.

Using \(a_n=a+(n-1)d\), \(a_7+a_{19}=(a+6d)+(a+18d)=2a_{13}\). So option A is always true; B has the wrong middle term. Exam tip: pair terms equidistant from the centre.

The nth term of an AP is \(a_n=a+(n-1)d\), so it is linear in \(n\). In \(n^2+2\), the consecutive differences are \(3,5,7\dots\), not constant. Exam tip: check whether the first differences remain equal.

The first term is \(a=13\), and the common difference is \(d=\frac{39}{2}-13=\frac{13}{2}\). The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{46}=13+45\times\frac{13}{2}=\frac{26+585}{2}=\frac{611}{2}\). Hence, option B is correct. A nearby option such as \(\frac{599}{2}\) can result from using an incorrect number of differences. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

For an AP, \(a_{58}-a_{22}=(58-22)d\). Thus, \(-468-0=36d\), so \(d=-13\). Now, from \(a_{22}=a_1+21d\), \(0=a_1+21(-13)\), giving \(a_1=273\). Hence, 273 is the correct option. If 263 were used, the 22nd term would not be 0. Exam tip: First find \(d\) by using the difference between the given term numbers.

For \(a_n=4n-7\), \(a_{n+1}-a_n=[4(n+1)-7]-(4n-7)=4\), so the common difference is constant and it is an AP. The differences in the other options vary with \(n\). Exam tip: in \(pn+q\), the common difference is \(p\).

For this AP, the first term is 310 and the common difference is \(d=291-310=-19\). Thus, each successive term decreases by 19. The term after 82 is \(82-19=63\). Since 82 is greater than 75 while 63 is less than 75, 63 is the first term below 75. Although 44 and 25 are also less than 75, they occur after 63. Exam tip: In a decreasing AP, compare the terms immediately on either side of the given boundary.

For an AP, \(a_n=a+(n-1)d\). Therefore, \(a_{35}-a_{11}=(35-11)d=24d\). Given \(a_{35}-a_{11}=288\), we get \(24d=288\), so \(d=288/24=12\). The gap between the term numbers is 24; using an incorrect gap gives a wrong value. Exam tip: use \(a_m-a_n=(m-n)d\) directly in such questions.

For option B, \(a_{n+1}-a_n=[5-3(n+1)]-(5-3n)=-3\), which is constant for every \(n\). Hence it is an AP. In option A, the difference depends on \(n\). Exam tip: test \(a_{n+1}-a_n\) for constancy.

The term indices increase by 8 each time. Hence, \(a_{15}-a_7=(15x-48)-(7x-8)=8x-40\), and \(a_{23}-a_{15}=8x-40\). Therefore, the next equal-step difference is also \(8x-40\). Thus, \(a_{31}=a_{23}+(8x-40)=23x-88+8x-40=31x-128\). The option \(31x-124\) results from an incorrect calculation of the constant term by 4. Exam tip: Terms separated by equal index gaps in an AP have equal differences.

Here, the first term is 52 and the common difference is 21. Thus, \(a_n=52+21(n-1)=21n+31\). For a term not exceeding 2000, \(21n+31\le 2000\), giving \(n\le 93\). Therefore, \(a_{93}=1984\) is the greatest term in the given range. The next term is 2005, which exceeds 2000. Exam tip: To find the greatest AP term below a limit, use the greatest integer value of n satisfying the inequality.

For \(a_n=5n-3\), the consecutive difference is \(a_{n+1}-a_n=5\), which is constant; hence it is an AP. In \(n^2+1\), the difference changes. Exam tip: any form \(pn+q\) represents an AP.

Let the first term be \(a\) and the common difference be \(d\). Then \(a_6+a_{13}=(a+5d)+(a+12d)=2a+17d=211\), while \(a_{22}=a+21d=256\). Doubling the second equation and subtracting the first gives \(25d=301\), so \(d=\frac{301}{25}\). Hence \(a=256-21d=\frac{79}{25}\). Therefore, \(a_{44}=a+43d=\frac{79}{25}+43\cdot\frac{301}{25}=\frac{13022}{25}\). The value \(539\) does not follow because the given data do not give \(d=14\). Exam tip: write \(a_n=a+(n-1)d\) carefully, especially when converting term indices.

In an AP, \(a_n=a+(n-1)d\). Thus, \(a_{12}=a+11d\) and \(a_{22}=a+21d\); their average is \(a+16d=a_{17}\). In option B, the midpoint of the indices is 16.5, not 17. Exam tip: a term midway between two indices equals the average of those terms.

Here (d=\frac{21}{2}). (a_{39}=-19+38\cdot\frac{21}{2}=380), so checking options is necessary.

(a_{30}=408), so (16d=357) and (d=\frac{357}{16}). (a_{46}=408+16\cdot\frac{357}{16}=765).