Finding the $n$th term of an AP

For an AP, \(a_{37}-a_{15}=(37-15)d\). Thus, \(-154-0=22d\), so \(d=-7\). Now, using \(a_{15}=a_1+14d\), we get \(0=a_1+14(-7)\), hence \(a_1=98\). If 84 were chosen, the 15th term would not be 0. Exam tip: Find \(d\) first by subtracting the two given terms.

The nth term of an AP has the form \(a_n=a+(n-1)d\), so it is linear in \(n\). In \(n^2+1\), consecutive differences are 3, 5, 7, ..., not constant. \(5\) is still an AP with \(d=0\). Exam tip: a squared or cubed \(n\) usually rules out an AP.

In \(a_n=a+(n-1)d\), the coefficient of \(n\) is the common difference \(d\). Here it is \(-3\), so every next term decreases by 3. Note that 7 is not the first term. Exam tip: identify \(d\) from the coefficient of \(n\).

For an AP, \(a_n=a+(n-1)d\). Therefore, \(a_{22}-a_7=(22-7)d=15d\). From the given relation, \(15d=120\), so \(d=8\). If the common difference were 9, the difference would be \(15\times 9=135\), not 120. Exam tip: when two terms are given, subtract their indices and multiply the result by \(d\).

In option A, \(a_{n+1}-a_n=[7-3(n+1)]-(7-3n)=-3\), which is constant for every \(n\); hence it is an AP. In option B, the difference changes. Exam tip: always test consecutive-term differences.

From the 4th to the 9th term and from the 9th to the 14th term, the gap is 5 terms each. \(a_9-a_4=(9x-16)-(4x-1)=5x-15\), so \(a_{19}-a_{14}\) is also \(5x-15\). Hence, \(a_{19}=(14x-31)+(5x-15)=19x-46\). The option \(19x-44\) has an incorrect addition of the constant terms. Exam tip: In an AP, equal gaps between term numbers give equal differences between the corresponding terms.

Here, the first term is 29 and the common difference is 12. Therefore, the nth term is \(a_n=29+12(n-1)\). For a term less than 800, \(29+12(n-1)<800\), which gives \(n-1<64.25\). Thus, the greatest possible integer value is \(n-1=64\), and the term is \(29+12\times64=797\). The next term, 809, is greater than 800. Exam tip: To find the greatest AP term below a limit, solve the inequality and take the greatest valid integer value of n.

For an AP, \(a_5=a+4d\). Thus, \(18=a+4\times3\) gives \(a=6\), so \(a_n=6+(n-1)3=3n+3\). In option B, the fifth term would be 21. Exam tip: find \(a\) before writing the general term.

Let the first term be \(a\) and the common difference be \(d\). Then \(a_3+a_8=(a+2d)+(a+7d)=2a+9d=76\), while \(a_{13}=a+12d=92\). Multiplying the second equation by 2 gives \(2a+24d=184\). Subtracting the first equation gives \(15d=108\), so \(d=7.2\). Hence \(a=92-12(7.2)=5.6\), and \(a_{27}=a+26d=5.6+26(7.2)=192.8\). The nearby distractor 196 does not satisfy the given conditions. Exam tip: first convert every stated AP term or sum into equations in \(a\) and \(d\).

Putting \(n=1\) gives \(a_1=12-4=8\). The coefficient of \(n\) is the common difference, so \(d=-4\); hence the AP decreases. Do not mistake the constant 12 for the first term; substitute \(n=1\).

The first term is \(a=-11\), and the common difference is \(d=\frac{-9}{2}-(-11)=\frac{13}{2}\). Therefore, \(a_{27}=a+(27-1)d=-11+26\times\frac{13}{2}=-11+169=158\). Hence, 158 is correct. \(169\) represents only \(26d\); the first term \(-11\) must also be added. Exam tip: while finding the \(n\)th term, use \(a_n=a+(n-1)d\), not \(a+nd\).

For option A, \(a_{n+1}-a_n=[4(n+1)-7]-(4n-7)=4\), which is constant, so it is an AP. For \(n^2-7\), the difference \(2n+1\) varies. Exam tip: \(pn+q\) form gives an AP.

In an AP, \(a_{19}-a_8=(19-8)d=11d\). Since \(a_{19}=a_8+99\), we get \(11d=99\), so \(d=9\). Now, \(a_{52}=a_8+(52-8)d=46+44\times9=46+396=442\). Therefore, the correct answer is \(442\). An option such as \(436\) results from not adding the correct number of common differences. Exam tip: when two terms are given, first use the difference in their indices to find \(d\).

Rewrite \(5-2n\) as \(3+(n-1)(-2)\), which matches \(a+(n-1)d\). Thus \(a=3\) and \(d=-2\). Option B wrongly treats the constant 5 as the first term. Exam tip: compare with the standard form first.

(d=-9). The last term greater than (-50) is (-40) if it appears, so always verify actual terms before choosing.

Putting \(n=1\), we get \(a_1=7-3=4\), so the first term is 4. The coefficient of \(n\) gives common difference \(-3\). Option A incorrectly treats 7 as the first term. Exam tip: find \(a_1\) first, then check the coefficient of \(n\).

From (218=36+(n-1)7), (182=7(n-1)), so (n=27). Daily practice time is the term, not the total time.

(197=8+(3n+1)9), so (189=27n+9) and (n=\frac{20}{3}). Read the index carefully before selecting an integer option.

In an AP, the difference between two terms equals the difference of their indices multiplied by the common difference. Thus, \(a_{5n}-a_n=(5n-n)d=4nd\). Hence \(205-49=4n\times13\), so \(156=52n\), giving \(n=3\). If \(n=4\), the difference would be \(208\), not 156. Exam tip: In such questions, subtract the given terms instead of first finding the initial term.

The nth term of an AP has the linear form \(a_n=pn+q\), where \(p\) is the common difference. In option A, consecutive terms differ by \(5\). In option B, the differences are not constant. Exam tip: identify the linear form first.