Finding the $n$th term of an AP

Given \(a_n=12n+q\), we get \(a_{8n}=12(8n)+q=96n+q\) and \(a_{3n}=12(3n)+q=36n+q\). Hence, \(a_{8n}-a_{3n}=(96n+q)-(36n+q)=60n\). Since \(60n=780\), \(n=13\). The constant \(q\) cancels because it occurs in both terms. Exam tip: substitute the required indices into the term formula first, then take their difference.

For an AP, \(a_k=a_1+(k-1)d\). As \(q-p=r-q\), we get \(p+r=2q\), so \(a_p+a_r=2a_q\). The product relation is associated with a GP, not an AP. Exam tip: equally spaced indices make the middle term the average.

In an AP, \(a_p-a_q=(p-q)d\). Hence, \(a_{3n+2}-a_{n+2}=[(3n+2)-(n+2)]d=2nd\). Therefore, \(148-52=2n\times12\), so \(96=24n\) and \(n=4\). If \(n=3\), the difference would be \(72\), not the given difference \(96\). Exam tip: When two terms are given, first find the difference between their subscripts.

In an AP, consecutive terms have a constant difference, so its nth term has the form \(a_n=a+(n-1)d\). Option A has \(a=7\) and \(d=4\). In option B, the difference changes. Exam tip: check whether \(n\) is linear for an AP.

In an AP, \(a_{r+6}-a_r=6d\). Thus, \(176-86=6d\), so \(d=15\). Now, \(a_{4r}-a_r=(4r-r)d=3rd\). Hence, \(536-86=3r\times15\), i.e. \(450=45r\), giving \(r=10\). If 9 were used, the difference between the terms would be \(3\times9\times15=405\), not 450. Exam tip: Subtracting two given AP terms eliminates the first term and directly relates the common difference to the difference in indices.