Finding the $n$th term of an AP

Given \(a_n=8n+q\), we get \(a_{6n}=8(6n)+q=48n+q\) and \(a_{2n}=8(2n)+q=16n+q\). Hence, \(a_{6n}-a_{2n}=32n\). Therefore, \(32n=384\) gives \(n=12\). The constant \(q\) cancels when the two terms are subtracted. Exam tip: substitute the index carefully into the given term formula before simplifying.

For \(a_n=3n-5\), \(a_{n+1}-a_n=[3(n+1)-5]-(3n-5)=3\), which is constant for every \(n\). Hence it is an AP. For \(n^2+1\), the difference changes. Exam tip: check consecutive-term differences.

In an AP, the difference between two terms equals the difference of their indices multiplied by the common difference. Thus, \(a_{2n+1}-a_{n+1}=[(2n+1)-(n+1)]d=nd\). Hence \(89-41=6n\), so \(48=6n\) and \(n=8\). If \(n=7\), the difference would be \(42\), not 48. Exam tip: first find the difference between the term indices; here it is \(n\).

The standard nth-term formula of an AP is \(a_n=a+(n-1)d\). Here \(a=7\) and \(d=5\), so \(a_n=7+5(n-1)\). In option A, substituting \(n=1\) gives 12, not 7. Exam tip: always check the \((n-1)\) factor.

The increase is zero for the first row, so the formula is \(a_n=24+4(n-1)\), not \(24+4n\). From \(24+4(n-1)=100\), we get \(n-1=19\), hence \(n=20\). The 19th row has only 96 seats. In exams, always check the \(n-1\) term.

Given \(a_n=7n+c\). Substituting \(n=6\), \(a_6=42+c=61\), so \(c=19\). Now \(a_{4r}=7(4r)+19=28r+19\). From \(28r+19=299\), we get \(28r=280\), hence \(r=10\). If 9 were used, the term would be 271, not 299. Exam tip: first find the constant \(c\) from the given term, then substitute the required index in the formula.

Using \(a_n=a+(n-1)d\), we get \((m-n)d=0\). Since \(m\ne n\), \(d=0\), so every term equals \(a\). Exam tip: first note that the indices are distinct.

In an AP, \(a_k=a+(k-1)d\). Thus \(a_{n-r}=a_n-rd\) and \(a_{n+r}=a_n+rd\), so their average is \(a_n\). Exam tip: remember this for equally spaced terms.

(d=\frac{144-46}{22-8}=7), so (a_{40}=144+18\times7=270). Moving from the nearer known term is faster.

(a_{3m}-a_m=2md=96), so (16m=96) and (m=6). In index questions, first check the position gap.

From (13k=104), (k=8). Therefore (a_{31}=8\times31+17=265). First find (k), then substitute the term number.

For \(a_n=5n-3\), \(a_{n+1}-a_n=5\), which is constant for every \(n\); hence it is an AP. In \(n^2+1\), successive differences vary. Exam tip: any linear form \(pn+q\) defines an AP.

\(a_{n+1}-a_n=[4(n+1)-7]-(4n-7)=4\). A sequence is an AP only when consecutive terms have a constant difference. A negative first term does not prove this. Exam tip: test \(a_{n+1}-a_n\).

(a_{12}=61) and (7d=42), so (d=6). (a_{26}=19+21\times6=145); the correct value is not in the listed options.

For \(a_n=7n-3\), \(a_{n+1}-a_n=[7(n+1)-3]-(7n-3)=7\), a constant difference; hence it is an AP. In \(n^2+7\), the differences vary. Exam tip: a linear expression in n represents an AP.

From (101=72+c), (c=29). (326=45r+29) does not give an integer (r), so the given data should be checked.

The terms are of the form (27+13(n-1)). The first term greater than (850) is (859) because the previous term is (846).

For option A, \(a_{n+1}-a_n=[5(n+1)-2]-(5n-2)=5\), which is constant for every \(n\); hence it is an AP. In option B, the difference changes. Exam tip: test consecutive differences first.

From the \(p\)th term to the \(n\)th term, the common difference is added \(n-p\) times, so \(a_n=a_p+(n-p)d\). Option C reverses the difference. Exam tip: count gaps between term positions, not the positions themselves.

In an AP, the difference between consecutive terms is constant. For \(a_n=7-3n\), \(a_{n+1}-a_n=-3\), so it is an AP. The \(n^2\) rule does not give a fixed difference. Exam tip: check whether the nth term is linear in n.