Finding the $n$th term of an AP

In an AP, the difference between two terms equals the difference of their indices multiplied by the common difference. Thus, \(a_{6n}-a_{2n}=(6n-2n)d=4n\times14=56n\). Given \(319-95=224\), we get \(56n=224\), so \(n=4\). Therefore, option 4 is correct. Exam tip: subtract the given terms to eliminate the first term; there is no need to find it separately.

For option A, \(a_{n+1}-a_n=\frac{5-2(n+1)}{3}-\frac{5-2n}{3}=-\frac{2}{3}\), which is constant; hence it is an AP. In B, the difference changes. Exam tip: an AP has an nth term linear in \(n\).

The difference between consecutive terms is \(8\), so the first term is \(a=u-9\) and the common difference is \(d=8\). Using \(a_n=a+(n-1)d\), \(295=(u-9)+28\times8=u+215\). Hence, \(u=80\). If \(u=78\), the 29th term would be \(293\), so it is not correct. Exam tip: for the \(n\)th term, use \(n-1\) common differences, not \(n\).

Given \(a_n=13n-6\), we have \(a_{5k}=13(5k)-6=65k-6\) and \(a_{2k}=13(2k)-6=26k-6\). Hence, \(a_{5k}-a_{2k}=(65k-6)-(26k-6)=39k\). Now \(39k=234\), so \(k=234/39=6\). If \(k=5\), the difference would be only \(195\), so it is not correct. Exam tip: substitute each index separately in the term formula before subtracting.

For \(a_n=7n-4\), \(a_{n+1}-a_n=7\), a constant non-zero value for every \(n\); hence it is an AP. Terms involving \(n^2\), \(2^n\), or \(1/n\) do not have constant consecutive differences. Exam tip: a linear form \(pn+q\) always represents an AP.

The positive multiples of 29 form the AP \(29, 58, 87, \ldots\), whose \(n\)th term is \(29n\). For the last term, \(29n<2500\). Since \(2500\div29\approx86.2\), the greatest integer value of \(n\) is 86. Hence the last term is \(29\times86=2494\). Although \(2523=29\times87\), it is greater than 2500. Exam tip: for “less than”, use the integer part of the quotient and verify the product.

The first term is \(a=920\), and the common difference is \(d=943-920=23\). The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{27}=920+(27-1)\times23=920+598=1518\). The value 1541 results from incorrectly using \(n\) instead of \((n-1)\). Exam tip: there are always \(n-1\) common differences between the first term and the \(n\)th term.

For an AP, \(a_n=a+(n-1)d\). Thus, \(a_7+a_{21}=2a+26d=224\) and \(a_{13}+a_{27}=2a+38d=368\). Subtracting the equations gives \(12d=144\), so \(d=12\). Then \(2a+26(12)=224\) gives \(a=-44\). Hence, \(a_{41}=a+40d=-44+40(12)=436\). Option 560 results from not accounting for the first term \(a\). Exam tip: in AP questions involving two sums of terms, subtract the equations first to find \(d\) quickly.

First, a₁₁ = 3(28) + 6 = 90. From a₆ to a₁₁ there are 5 equal steps, so 5d = 90 − 28 = 62. Thus a₃₁ = a₆ + 25d = 28 + 25(62/5) = 338.

Here, the first term is \(a=42\) and the common difference is \(d=13\). The \(n\)th term is \(a_n=42+13(n-1)\). From \(42+13(n-1)<1000\), we get \(13n<971\), so \(n<74.69\). Hence, there are \(74\) terms less than 1000. The \(75\)th term is \(1017\), which is not less than 1000. Exam tip: after solving an inequality for the number of terms, take the greatest integer satisfying it.

The \(n\)th term is obtained by adding the common difference \(d\), \((n-1)\) times to the first term \(a\); hence \(a_n=a+(n-1)d\). Using \(a+nd\) adds one extra \(d\) even for the first term. Exam tip: put \(n=1\); the result must be \(a\).

Using aₙ = a₅ + (n − 5)d, we get a₁₂ + a₁₉ = (31 + 7d) + (31 + 14d) = 242. Hence 21d = 180, so d = 60/7. Therefore, a₃₃ = 31 + 28(60/7) = 271.

Here, the first term is \(a=-47\) and the common difference is \(d=16\). Thus, \(a_n=-47+16(n-1)=16n-63\). For \(a_n>500\), \(16n-63>500\), so \(n>35.1875\). The smallest integer value is \(n=36\), giving \(a_{36}=513\). Option 497 is not greater than 500. Exam tip: For “greater than,” use the next integer term satisfying the inequality.

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.

The common difference is d = (144 − 46) ÷ (22 − 8) = 7. From the 22nd term to the 40th term there are 18 steps, so a₄₀ = 144 + 18 × 7 = 270.

The direct answer is B: \(m=6\). In an AP, the term at position \(r\) is \(a_r=a+(r-1)d\). Instead of finding the first term, subtract the two given terms. The positions are \(3m\) and \(m\), so their difference is \(2m\). Thus \(a_{3m}-a_m=(3m-m)d=2m\times8\). The values give \(152-56=96\), so \(16m=96\), and dividing by 16 gives \(m=6\). Option A, 5, would give a difference of 80, not 96. Option B, 6, satisfies the equation. Option C, 7, would give 112, and option D, 8, would give 128; both are too large. The key idea is that AP terms change by \(d\) for every one-position increase, so multiply the index gap by \(d\).

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 \(a_{n+1}-a_n\).