Finding the $n$th term of an AP

Let the first term be \(a\) and common difference be \(d\). Then \(a_{10}+a_{30}=2a+38d=500\) and \(a_{18}+a_{38}=2a+54d=820\). Subtracting gives \(16d=320\), so \(d=20\). From \(2a+38(20)=500\), we get \(a=-130\). Hence, \(a_{58}=a+57d=-130+57(20)=1010\). The nearby option 1030 can result from incorrectly using \(58d\); remember that \(a_{58}=a+57d\). Exam tip: use \(a_n=a+(n-1)d\).

For \(a_n=5-2n\), \(a_{n+1}-a_n=[5-2(n+1)]-(5-2n)=-2\), which is constant for every n. Hence it is an AP. The differences for \(3n^2-2\) are not constant. Exam tip: in \(a_n=pn+q\), the common difference is \(p\).

Here, the first term is \(a=75\) and the common difference is \(d=98-75=23\). The \(n\)th term is \(a_n=75+23(n-1)\). Using \(75+23(n-1)<2500\), we get \(n-1<\frac{2425}{23}\approx105.43\). Hence, the greatest integer value of \(n\) is \(106\). Therefore, 106 terms are less than 2500. For \(n=107\), the term is \(2513\), which is not less than 2500. Exam tip: retain the strict inequality \((<)\) when the question says “less than.”

The standard AP form is \(a_n=a+(n-1)d\). Here, \(a_1=7(1)-3=4\), and each successive term increases by 7, so \(d=7\). The \(-3\) is only a constant term, not the common difference. Exam tip: find \(a_1\) and \(a_2\), then subtract.

Let d be the common difference. Since a₁₆ = a₇ + 9d and a₂₅ = a₇ + 18d, their sum gives 116 + 27d = 638, so d = 522/27 = 58/3. Therefore, a₄₉ = a₇ + 42d = 58 + 42 × 58/3 = 870.

Here, the first term is \(a=-92\) and the common difference is \(d=23\). Thus, \(a_n=a+(n-1)d=-92+23(n-1)=23n-115\). From \(23n-115>900\), we get \(n>44.13\), so the smallest integer value is \(n=45\). Hence, \(a_{45}=920\), which is the first term greater than 900. The close distractor 897 is the 44th term, so it is still less than 900. Exam tip: for the first term greater than a number, solve the inequality and take the next integer value of \(n\).

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.