Finding the $n$th term of an AP

Correct answer: C. 8

The difference between these terms is a_{2p} − a_p = (2p − p)d = 6p. Since 74 − 26 = 48, we get 6p = 48, so p = 8.

Correct answer: B. 185

The common difference is d = (95 − 23)/(18 − 6) = 72/12 = 6. From the 18th term to the 33rd term there are 15 steps, so a_33 = 95 + 15 × 6 = 185.

Correct answer: B. (-101)

The direct answer is option B, -101. The sequence is an arithmetic progression: 53, 46, 39, ... Each term decreases by 7, so the common difference is d=-7. We need the first term that is less than -100, not merely equal to or close to it. Continue the sequence near the target: after 53 and repeated subtraction of 7, the terms are ...,-87,-94,-101,-108. The term -94 is not less than -100 because -94 is greater than -100. The next term is -101, and -101<-100, so it is the first qualifying term. Option A, -94, is wrong because it has not crossed below -100. Option B, -101, is correct because it is below -100 and comes immediately after -94. Option C, -108, is also less than -100, but it is not the first such term; -101 occurs earlier. Option D, -115, is likewise below -100 but occurs even later. Algebraically, the nth term is a_n=53+(n-1)(-7)=60-7n. Solving 60-7n<-100 gives -7n<-160; reversing the sign when dividing by -7 gives n>160/7≈22.86, so the first integer n is 23. Then a_23=60-161=-101. The key memory point is that with negative numbers, -101 is smaller than -100, while -94 is larger.

Correct answer: C. \(x+181\)

There are 12 term-intervals between the 8th and the 20th terms. Hence, \(12d=(x+91)-(x+19)=72\), so \(d=6\). From the 20th term to the 35th term, there are 15 intervals; therefore, \(a_{35}=x+91+15\times6=x+181\). Thus, \(x+181\) is correct. \(x+175\) would result from incorrectly using only 14 intervals. Exam tip: first find \(d\) from the two given terms, then count the term-intervals from the nearer known term.

Correct answer: D. 158

First, a_10 = 4 × 14 + 6 = 62. The 10th and 4th terms are six positions apart, so 62 − 14 = 6d, giving d = 8. Thus a_22 = a_10 + 12d = 62 + 12 × 8 = 158.

Correct answer: A. \(a_n=5n-3\)

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

Correct answer: C. 10.5

Using a_6 = 61 gives 42 + c = 61, so c = 19. Then a_{4r} = 28r + 19 = 313, so 28r = 294 and r = 10.5. This gives the integer term index 4r = 42.

Correct answer: A. मध्य पद

In an AP with an odd number of terms, the middle term is the arithmetic mean of the first and last terms. Hence, first term + last term = 2 × middle term. Exam tip: first check whether the number of terms is odd.

Correct answer: A. First term 4 and common difference \(-3\)

The standard AP form is \(a_n=a+(n-1)d\). Here, \(a_1=7-3(1)=4\), and each successive term changes by \(-3\); hence \(d=-3\). Exam tip: substitute \(n=1\) to verify the first term.

Correct answer: A. It is a decreasing AP with common difference \(-3\)

\(a_{n+1}-a_n=[5-3(n+1)]-(5-3n)=-3\), which is constant for every \(n\). Hence it is an AP; the negative difference makes it decreasing. Exam tip: the coefficient of \(n\) gives the common difference.

Correct answer: B. \(\frac{349}{2}\)

The first term is \(a=8\), and the common difference is \(d=\frac{25}{2}-8=\frac{9}{2}\). Therefore, \(a_{38}=a+(38-1)d=8+37\times\frac{9}{2}=\frac{16+333}{2}=\frac{349}{2}\). Hence, option B is correct. Choosing \(\frac{341}{2}\) results from an error in counting the number of differences or calculating \(d\). Exam tip: for the \(n\)th term, always use \(a_n=a+(n-1)d\), not \(a+nd\).

Correct answer: B. \(136\)

For an AP, \(a_{42}-a_{18}=(42-18)d\). Thus, \(-192-0=24d\), so \(d=-8\). Using \(a_{18}=a_1+17d\), we get \(0=a_1+17(-8)\), hence \(a_1=136\). If \(128\) were used, the 18th term would not be 0. Exam tip: Find \(d\) first by subtracting the two given terms.

Correct answer: B. \(a_n=20-3n\)

An AP has nth term \(a_n=a+(n-1)d\). In \(20-3n\), the coefficient of n is \(-3\), so consecutive terms differ by \(-3\). For \(20-3n^2\), the difference is not constant. Exam tip: check the coefficient of n in a linear expression.

Correct answer: B. 34

The common difference is \(d=154-169=-15\). The terms around the boundary are \(49, 34, 19,\ldots\). Since \(49\) is not less than \(40\), but the very next term \(34\) is less than \(40\), the first required term is \(34\). Exam tip: In a decreasing AP, write the consecutive terms around the given boundary to identify the first qualifying term.

Correct answer: C. 9

In an AP, the difference between two terms equals the difference of their positions multiplied by the common difference: \(a_{27}-a_8=(27-8)d=19d\). Given \(a_{27}-a_8=171\), we get \(19d=171\), so \(d=9\). Hence, option C is correct. If the common difference were 8, the term difference would be \(19\times8=152\), not 171. Exam tip: use \(a_m-a_n=(m-n)d\) to solve such questions quickly.

Correct answer: C. \(23x-56\)

The terms are six positions apart, so their differences must be equal. \(a_{11}-a_5=(11x-20)-(5x-2)=6x-18\), and \(a_{17}-a_{11}=6x-18\). Hence, \(a_{23}=a_{17}+(6x-18)=(17x-38)+(6x-18)=23x-56\). Therefore, option C is correct. Exam tip: When equally spaced AP terms are given, use the equal difference between those terms to find the next one.

Correct answer: C. \(997\)

Here, the first term is \(a=31\) and the common difference is \(d=14\). Thus, \(a_n=31+14(n-1)\). To find the greatest term below \(1000\), use \(31+14(n-1)<1000\), which gives \(n-1\leq69\). Therefore, the term is \(31+14\times69=997\). The next term is \(1011\), which exceeds \(1000\). Exam tip: for the greatest AP term below a limit, use an inequality and take the greatest possible integer value of \(n-1\).

Correct answer: A. \(p\)

The common difference is \(a_{n+1}-a_n\). Here, \(a_{n+1}=p(n+1)+q=pn+p+q\), so the difference is \(p\). The constant \(q\) only shifts every term. Exam tip: identify the coefficient of \(n\).

Correct answer: B. \(\frac{5288}{19}\)

Let the first term be \(a\) and the common difference be \(d\). Then \(a_4+a_9=(a+3d)+(a+8d)=2a+11d=95\), while \(a_{16}=a+15d=137\). Doubling the second equation and subtracting the first gives \(19d=179\), so \(d=\frac{179}{19}\). Therefore, \(a_{31}=a_{16}+15d=137+15\left(\frac{179}{19}\right)=\frac{5288}{19}\). The value \(263\) would require \(d=9\), which does not satisfy both given conditions. Exam tip: Use \(a_{31}=a_{16}+15d\) to calculate a distant term efficiently.

Correct answer: D. 332

First, a₁₂ = 2(37) + 22 = 96. The common difference is d = (96 − 37)/(12 − 8) = 59/4. Thus, a₂₈ = a₈ + 20d = 37 + 20(59/4) = 332.

Correct answer: B. 212

The common difference is d = −11/2 − (−13) = 15/2. Using aₙ = a₁ + (n − 1)d, a₃₁ = −13 + 30(15/2) = −13 + 225 = 212.

Correct answer: A. \(d=0\)

For an AP, \(a_n=a+(n-1)d\). If \(d=0\), then \(a_n=a\), so every term is constant. Setting only \(a=0\) does not make terms constant. Exam tip: the coefficient of \(n\) is \(d\).

Correct answer: C. 582

In an AP, the difference between two terms equals the difference in their positions multiplied by the common difference. Thus, \(a_{21}-a_9=12d=132\), so \(d=11\). Now, \(a_{57}=a_9+(57-9)d=54+48\times11=582\). Therefore, option C is correct. The value 570 does not include the full increase of 48 common differences. Exam tip: When moving from one known term to another, use the difference between their term numbers.

Correct answer: A. \(a_n=5n-3\)

The nth term of an AP is linear: \(a+(n-1)d\). For \(5n-3\), \(a_{n+1}-a_n=5\), a constant difference. For \(n^2+1\), the difference is \(2n+1\), which varies. Exam tip: check consecutive-term differences.

Correct answer: A. (-71)

Here, the first term is \(a=121\) and the common difference is \(d=-12\). Thus, \(a_n=121-12(n-1)\). We get \(a_{17}=-71\), and the next term is \(a_{18}=-83\). Since \(-71>-75\) but \(-83<-75\), the last term greater than \(-75\) is \((-71)\). \((-83)\) is a close distractor, but it is less than \(-75\). Exam tip: In a decreasing AP, check the two consecutive terms on either side of the given limit.