Finding the $n$th term of an AP

Putting (n=1), (a_1=7), and consecutive terms differ by (2). In a direct formula, (d) is the coefficient of (n).

For an arithmetic progression, the nth term is aₙ = a + (n − 1)d. We need aₙ = 0, with a = 30 and d = −2. Substitution gives 0 = 30 + (n − 1)(−2). Rearranging, 2(n − 1) = 30, so n − 1 = 15 and n = 16. Therefore the zero term is the 16th term, so option C is correct. The negative common difference means the terms decrease by 2: the 14th term is 4, the 15th is 2, and the 16th is 0. Option B stops one step early, while option D goes one step too far and gives −2. This direct check confirms both the index and the sign.

Here, the first term is 1 and the common difference is 3, so the nth term is \(a_n=1+3(n-1)=3n-2\). From \(3n-2<100\), we get \(n<34\); hence the greatest integral value of n is \(33\). Therefore, \(a_{33}=97\). Although 99 is less than 100, it is not a term of this AP because the terms increase by 3 from 1. Exam tip: write \(a_n<\) the given number and find the greatest possible integer value of n.

For an AP, \(a_n=a_1+(n-1)d\). Given \(a_{11}=52\) and \(d=5\): \(52=a_1+10\cdot 5=a_1+50\Rightarrow a_1=2\). Then \(a_2=a_1+d=2+5=7\).

This is an AP with (a=10), (d=3). (a_{12}=10+11\times3=43) cm. In word problems, identify the first term and difference.

The direct answer is B, ₹900. The weekly deposits form an arithmetic progression: the first deposit is \(a=₹200\), the common increase is \(d=₹50\), and the required term is the 15th term. Use \(a_n=a+(n-1)d\): \(a_{15}=200+(15-1)50=200+14\times50=200+700=₹900\). Option A, ₹850, is one weekly increase too small. Option B, ₹900, correctly gives the deposit made in week 15. Option C, ₹950, adds 15 increases instead of 14, forgetting that the first week already starts at ₹200. Option D, ₹1000, adds still more than the required increases. The question asks for one week’s deposit, not the total saved over fifteen weeks. The exam cue is to count the increases between week 1 and week n as \(n-1\), not n.

This forms an arithmetic progression with first term \(a=18\), common difference \(d=2\), and \(n=20\). Using \(a_n=a+(n-1)d\), \(a_{20}=18+(20-1)\times2=18+38=56\). Therefore, the 20th row has 56 seats. Choosing 58 would result from incorrectly adding the difference one extra time. Exam tip: in the \(n\)th term, the common difference is added \(n-1\) times.

This is an arithmetic progression with first term \(a=25\) cm and common difference \(d=1.5\) cm. The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{30}=25+(30-1)\times1.5=25+43.5=68.5\) cm. Hence, 68.5 cm is correct. The value 70 cm can result from incorrectly adding \(30\times1.5\); the plant grows only 29 times after the first day. Exam tip: use \((n-1)d\), not \(nd\), for the \(n\)th term of an AP.

The direct answer is option A, 240 items. This is a decreasing arithmetic progression because the production changes by the same amount every hour. The first-hour production is a=500, and production decreases by 20, so the common difference is d=-20, not +20. The nth-term formula is a_n=a+(n-1)d. For the fourteenth hour, a_14=500+(14-1)(-20)=500+13(-20). First calculate 13×20=260; then 500-260=240. Thus the machine makes 240 items in the fourteenth hour. Option A, 240, matches the calculation. Option B, 260, would result from subtracting 12 groups of 20, which corresponds to the thirteenth hour, so it is one hour early. Option C, 280, corresponds to subtracting 11 groups and is even earlier. Option D, 300, corresponds to subtracting 10 groups and is not the fourteenth-hour value. The factor is 13, not 14, because the first hour already has the starting value 500; only the changes after the first hour are counted. In a decreasing AP, write d as negative. A quick check is to continue: hour 1 is 500, hour 2 is 480, and after thirteen decreases of 20 the answer is 240.

This is an arithmetic progression with first term \(a=12\) and common difference \(d=16-12=4\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{18}=12+(18-1)\times4=12+68=80\). Hence, the 18th chapter will have 80 pages. Option 84 would result from incorrectly using a larger increase than the actual common difference. Exam tip: Always use \(n-1\), not \(n\), in the nth-term formula.

The direct answer is B, 54. We are given the general term \(a_n=9n+2\). Substitute the required positions separately: \(a_{11}=9(11)+2=99+2=101\), and \(a_5=9(5)+2=45+2=47\). Therefore \(a_{11}-a_5=101-47=54\). We can also see that the coefficient of n is the common difference, so \(d=9\); the positions differ by \(11-5=6\), giving \(6d=6\times9=54\). Option A, 45, is only part of the calculation and equals \(9\times5\). Option B, 54, is correct. Option C, 63, uses an incorrect position difference or extra amount. Option D, 72, also overcounts the difference. Always substitute the complete expression, including the constant +2, or use the position difference carefully.

In an AP, the difference between two terms equals the corresponding number of common differences. Thus, \(a_{25}-a_{15}=10d\), so \(120-70=10d\) and \(d=5\). Now, using \(a_{15}=a_1+14d\), we get \(70=a_1+14\times5\), hence \(a_1=0\). Option 5 is the common difference, not the first term. Exam tip: In \(a_n=a_1+(n-1)d\), remember to use \(n-1\).

Here, the first term is \(a=x\) and the common difference is \(d=4\). Using \(a_n=a+(n-1)d\), we get \(50=x+(10-1)\times4=x+36\). Hence, \(x=14\). If 16 were chosen, the 10th term would be \(16+36=52\), not 50. Exam tip: In the nth-term formula, use \(n-1\), not \(n\).

The direct answer is option C, 20. An arithmetic progression changes by the same amount each time. Here the first term is a=27 and the common difference is d=24-27=-3. The nth-term rule is a_n=a+(n-1)d. Therefore, -30=27+(n-1)(-3). Subtract 27: -57=-3(n-1). Dividing by -3 gives 19=n-1, so n=20. Check: the 20th term is 27+19(-3)=27-57=-30. Option A, 18, would give 27+17(-3)=-24, so it is wrong. Option B, 19, gives -27, so it is wrong. Option C gives exactly -30. Option D, 21, gives -33, so it is also wrong. The minus sign in the common difference must be retained. Memory cue: for the nth term, always use n-1 differences, not n differences.

(d=\frac{64-29}{14-7}=5), hence (a_{21}=64+7\times5=99). Equal position gap (7) gives term gap (35).

The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{13}=4+(13-1)\times\frac{3}{2}=4+12\times\frac{3}{2}=4+18=22\). Hence, 22 is correct. Getting 21 would mean the common difference has not been multiplied correctly. Exam tip: always use \((n-1)\), not \(n\), in the formula for \(a_n\).

Given \(a_n=50-4n\). For the term whose value is \(-10\), set \(50-4n=-10\). Thus, \(4n=60\), so \(n=15\). Therefore, \(-10\) is the 15th term of the AP. The 14th term is \(50-4(14)=-6\), so it is not correct. Exam tip: To find a term number, equate the given term value to \(a_n\) and solve for \(n\).

The general term of an arithmetic progression is \(a_n=a+(n-1)d\), and consecutive terms differ by the constant common difference d. The indices 10 and 18 are eight places apart. Their term values differ by 55−31=24, so \(8d=24\), giving \(d=3\). Moving from the tenth term back to the sixth term means moving four positions backward.

Each backward move subtracts the common difference. Therefore, \(a_6=a_{10}-4d=31-4(3)=31-12=19\). Choice B is correct. A common error is to use five steps instead of four; the index difference is \(10−6=4\), not the value of either index. The result is also consistent because adding 3 twelve times from the sixth term reaches the eighteenth term.

The first term is 8 and the common difference is 6. Using \(a_n=a+(n-1)d\), \(74=8+(n-1)\times6\). Thus, \(66=6(n-1)\), so \(n-1=11\) and \(n=12\). If \(n=13\), the term would be \(8+12\times6=80\), not 74. Exam tip: after finding \(n-1\), remember to add 1 to obtain the term number.

Subtracting the relations gives (d=-1), and substitution gives (a_{p+q}=0). Even in symbolic APs, use (a_n=a+(n-1)d).

Given \(a_3=8\) and \(a_8=3\), we have \(a_8-a_3=5d\). Thus, \(3-8=5d\), so \(d=-1\). Now \(a_{11}=a_8+3d=3+3(-1)=0\). Hence, 0 is the correct answer. Option 1 is the value of \(a_{10}\), obtained by counting one term less. Exam tip: When using two known terms of an AP, take the difference of their term numbers to find the number of common differences.

In this AP, (a=105), (d=7), so (a_{50}=105+49\times7=448). In an AP of multiples, choose the first correct multiple.

In (9,18,27,\ldots), (9n<200), so the greatest (n=22) and the term is (198). For multiples, take the largest integer below the limit.

The given formula is \(a_n=6n+5\). Putting \(n=32\), \(a_{32}=6\times32+5=192+5=197\). Hence, 197 is correct. The value 191 would result from incorrectly using \(n=31\). Exam tip: Substitute the required term number directly for \(n\) in the formula.

In an arithmetic progression, the difference between consecutive terms is constant and equals the common difference. Since a_(n+1) − a_n = 8, we have d = 8. Starting from the known fourth term, moving to the sixteenth term requires 16 − 4 = 12 equal steps. Each step adds 8, so a_16 = a_4 + 12d = 30 + 12 × 8 = 30 + 96 = 126. Thus option C is correct. A common mistake is to use 16 steps instead of the difference in indices, or to add only 11 differences. Option B corresponds to adding 92, and option D corresponds to adding 100; neither follows the constant-difference rule. The result can also be checked by writing a_n = 30 + (n − 4)8.