Finding the $n$th term of an AP

In an AP, \(a_1=a\) and \(a_2=a+d\). Hence, \(a_1+a_2=a+(a+5)=33\), so \(2a=28\) and \(a=14\). Therefore, \(a_{12}=a+11d=14+11\times5=69\). Choosing 71 would result from using the number of terms or the common difference incorrectly. Exam tip: while finding the \(n\)th term, use \(a_n=a+(n-1)d\), ensuring that the multiplier is \(n-1\).

(a_{10}) is the middle term between (a_6) and (a_{14}), so (a_{10}=\frac{100}{2}=50). The average of equally spaced terms is the middle term.

Here, the first term is \(a=90\) and the common difference is \(d=-7\). The terms proceed as \(90,83,76,\ldots,27,20\). The next term after \(27\) is \(20\), which is not greater than \(20\). Therefore, the last term greater than \(20\) is \(27\). Exam tip: In a decreasing AP, check the two consecutive terms around the given limit.

The first term is 6 and the common difference is 14 − 6 = 8. Using the nth-term formula, aₙ = a₁ + (n − 1)d, we get a₂₃ = 6 + 22 × 8 = 182.

The nth term of an arithmetic progression is given by aₙ = a₁ + (n − 1)d. Here the first term is a₁ = 84, and the common difference is d = 78 − 84 = −6. The negative sign shows that the sequence decreases by 6 at every step. Substituting n = 18 gives a₁₈ = 84 + (18 − 1)(−6) = 84 − 17 × 6 = 84 − 102 = −18. Therefore option B is correct. The factor 17 is essential because the first term requires zero increments, and the eighteenth term is reached after seventeen transitions. Using a positive difference would reverse the trend, while using 18 differences would count one step too many and lead to a different value.

The governing concept is the nth-term formula for an arithmetic progression: aₙ = a + (n − 1)d. Here the first term is a = 17, the common difference is d = 9, and the required term is 170. Substitute these values: 170 = 17 + (n − 1)9. Subtracting 17 gives 153 = 9(n − 1). Dividing by 9 gives n − 1 = 17, so n = 18. Therefore option C is correct. The expression n − 1 counts the number of equal jumps after the first term; seventeen jumps take us from the first term to the eighteenth term. Option B, 17, results from forgetting to add 1 after finding the number of jumps. Options A and D similarly use an incorrect count of intervals.

For an AP, \(a_n=a+(n-1)d\). Thus, \(71=a+11\times5\), so \(a=71-55=16\). If 18 were taken as the first term, the 12th term would be \(18+11\times5=73\), not 71. Exam tip: use \((n-1)\), not \(n\), in the formula for the nth term.

This is an nth-term and term-position problem in an arithmetic progression. The first term is a = 9, and the common difference is d = 17 − 9 = 8. Let 201 be the nth term. Using aₙ = a + (n − 1)d, we obtain 201 = 9 + 8(n − 1). Subtract 9 from both sides: 192 = 8(n − 1). Dividing by 8 gives n − 1 = 24, and therefore n = 25. Thus 201 is the 25th term, so option C is correct. The integer value confirms that 201 belongs to the progression. A useful check is that 24 increments of 8 added to the first term give 9 + 24 × 8 = 201. The nearby alternatives arise from counting increments as terms or taking one extra increment.

The first term is −28 and the common difference is −20 − (−28) = 8. Thus, a₂₁ = −28 + (21 − 1) × 8 = −28 + 160 = 132.

In an AP, the difference between two terms equals the difference in their positions multiplied by the common difference. Thus, \(a_{16}-a_7=(16-7)d\). Hence, \(90-36=9d\), so \(d=6\). If the difference were 7, the difference across 9 positions would be \(63\), not the given \(54\). Exam tip: When two terms of an AP are known, use \(d=\frac{a_n-a_m}{n-m}\) directly.

The 9 steps from the 6th to the 15th term increase the value by 101 − 38 = 63, so the common difference is 63 ÷ 9 = 7. From the 15th to the 24th term is another 9 steps; therefore, a₂₄ = 101 + 9 × 7 = 164.

Here, the first term is \(a=120\) and the common difference is \(d=109-120=-11\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Thus, \(a_{13}=120+(13-1)(-11)=120-132=-12\). Therefore, \(-12\) is correct. \(-10\) can result from incorrectly counting the number of differences or using \(n\) instead of \((n-1)\). Exam tip: for the \(n\)th term, always use \((n-1)\) common differences.

For an AP, \(a_n=a+(n-1)d\). Substituting \(a=31\), \(n=19\), and \(a_{19}=139\), we get \(139=31+18d\). Thus, \(108=18d\), so \(d=6\). If the common difference were 5, the 19th term would be \(31+18\times5=121\), not 139. Exam tip: the common difference is added \(n-1\) times to reach the \(n\)th term.

To find a term from a known term of an AP, use \(a_n=a_m+(n-m)d\). Thus, \(a_{19}=a_5+(19-5)d=24+14\times8=24+112=136\). Hence, 136 is correct. Getting 144 would indicate that the difference between the term positions was taken incorrectly. Exam tip: the number of common differences between \(a_m\) and \(a_n\) is always \(n-m\).

The first term is a = 2.4 and the common difference is d = 4.0 − 2.4 = 1.6. Using aₙ = a + (n − 1)d, a₁₆ = 2.4 + 15 × 1.6 = 26.4.

Given \(a_n=95-5n\), substitute \(n=26\): \(a_{26}=95-5\times26=95-130=-35\). Hence, option A is correct. \(-30\) would result from an error in multiplication or subtraction. Exam tip: after substituting \(n\), multiply first and then subtract.

The first term is a = 2/3 and the common difference is d = 2/3. Thus, a₂₈ = a + 27d = 2/3 + 27 × 2/3 = 56/3.

For an AP, \(a_n=a_1+(n-1)d\). Thus, \(92=a_1+(14-1)\times7=a_1+91\), so \(a_1=1\). Option \(85\) would result from subtracting \(d\) only once, but moving from the 14th term to the first term requires subtracting \(13d\). Exam tip: always use \(n-1\) in the formula for the \(n\)th term.

The governing idea is the nth-term formula of an arithmetic progression. The first term is a = 26 and the common difference is d = 35 − 26 = 9. If 206 is the nth term, then aₙ = a + (n − 1)d gives 206 = 26 + 9(n − 1). Subtracting 26 yields 180 = 9(n − 1). On division by 9, n − 1 = 20, so n = 21. Therefore 206 is the 21st term and option C is correct. The result can be checked directly: starting with 26, adding 9 twenty times gives 26 + 180 = 206; the initial term has index 1, so twenty increases lead to index 21. Options B and D correspond to one fewer or one more increase, while option A is even farther from the correct position.

The first term is a = −45 and the common difference is d = 9. Therefore, a₂₀ = a + 19d = −45 + 19 × 9 = 126.

There are 17 − 9 = 8 steps between a₉ and a₁₇, so the common difference is d = (108 − 52)/8 = 7. There are also 25 − 17 = 8 steps from a₁₇ to a₂₅, so a₂₅ = 108 + 8 × 7 = 164.

For an AP, \(a_n=a+(n-1)d\). Here, \(a=22\), \(n=32\), and \(a_{32}=177\). Thus, \(177=22+31d\), so \(31d=155\) and \(d=5\). If the common difference were 4, the 32nd term would be \(22+31\times4=146\), not 177. Exam tip: the common difference is added \(n-1\) times to obtain the \(n\)th term.

Here, the first term is \(a=105\) and the common difference is \(d=98-105=-7\). Therefore, \(a_n=a+(n-1)d=105-7(n-1)=112-7n\). For a negative term, \(112-7n<0\), so \(n>16\). The 16th term is \(0\), which is not negative; hence the 17th term, \(-7\), is the first negative term. Exam tip: For a negative term use \(<0\), not \(\leq0\).

The common difference is 8, so the terms are 13 + 8(n − 1). The term 245 occurs when n = 30; the next term is 253, which is greater than 250. Thus, 245 is the greatest term below 250.

The first term is a₁ = 4(1) + 11 = 15. The common difference is aₙ₊₁ − aₙ = 4, since increasing n by 1 increases the expression by 4.