Arithmetic Progression

Here, the first term is \(a=14\), the common difference is \(d=22-14=8\), and \(n=15\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), \(S_{15}=\frac{15}{2}[2(14)+14(8)]=\frac{15}{2}(140)=1050\). Hence, 1050 is correct. A value such as 1040 usually results from an addition or multiplication error. Exam tip: calculate \(n-1\) first before substituting in the formula.

The governing relation is a_n = a_1 + (n - 1)d. Applying it to the two given terms gives a_5 = a_1 + 4d = 33 and a_11 = a_1 + 10d = 75. Subtracting the first equation from the second eliminates a_1: 6d = 42, so d = 7. Substituting into a_1 + 4(7) = 33 gives a_1 + 28 = 33, hence a_1 = 5. Option B is correct. A quick check produces the fifth term 5 + 4×7 = 33 and the eleventh term 5 + 10×7 = 75. The other options fail one or both of these conditions.

Here, the first term is \(a=7\) and the common difference is \(d=5\). The \(n\)th term is \(a_n=a+(n-1)d\). So, \(72=7+(n-1)\times5\), giving \(65=5(n-1)\), hence \(n-1=13\) and \(n=14\). Therefore, 72 is the 14th term of the AP. The 13th term is 67, so 13 is not correct. Exam tip: When the last term is given, equate it to \(a_n\) and solve for \(n\).

Here, \(S_4\) denotes the sum of the first four terms: \(9+17+25+33=84\). Therefore, the correct answer is 84. The value 80 is obtained by adding only the first and last terms, so it is not the required sum. Exam tip: when only a few terms are given, direct addition is quick and reliable.

Here, the first term is \(a=18\) and the common difference is \(d=25-18=7\). The \(n\)th term is \(a_n=a+(n-1)d\). So, \(18+(n-1)\times7=123\), giving \((n-1)\times7=105\). Hence \(n-1=15\) and \(n=16\). Therefore, 123 is the 16th term. The 15th term is \(18+14\times7=116\), so it is not correct. Exam tip: after finding \(n-1\), remember to add 1 to get the term number.

In an AP, consecutive differences are equal: \(q-p=r-q\). Rearranging gives \(2q=p+r\), so the middle term is the average of the outer terms. \(q^2=pr\) is the corresponding condition for a GP. Exam tip: check the average-middle-term property quickly.

(S_{14}=\frac{14}{2}[12+13\times7]=721). In exams, calculate the bracket and multiplication carefully.

There are (6) differences from the sixth to the twelfth term, so (a_{12}=47+6\times9=101). In exams, count the difference in positions.

Here, the first term is \(a=84\) and the common difference is \(d=75-84=-9\). Using \(a_n=a+(n-1)d\), we get \(12=84+(n-1)(-9)\). Thus, \(n-1=8\), so \(n=9\). Therefore, 12 is the ninth term. The tenth term is 3, so the closest distractor is not correct. Exam tip: always use a negative common difference for a decreasing AP.

From (216=\frac{8}{2}[12+7d]), (d=6). In exams, keep (2a) and ((n-1)d) correct in the (S_n) formula.

Here, the first term is \(a=21\) and the common difference is \(d=28-21=7\). Using \(a_n=a+(n-1)d\), we get \(a_9=21+8\times7=77\) and \(a_{13}=21+12\times7=105\). Therefore, \(a_9+a_{13}=77+105=182\). Option 168 may result from using an incorrect difference instead of \((n-1)d\). Exam tip: while finding the \(n\)th term of an AP, always use \((n-1)d\).

Given a_n=74-5n and a_n=19. So, 74-5n=19, which gives 5n=55 and n=11. Hence, 19 is the 11th term. For example, the 12th term is 74-5(12)=14, not 19. Exam tip: To find a term number, equate the given term value to a_n and solve for n.

This is the sum of the first (12) multiples of (8), which is (624). In exams, identify sequences of multiples quickly.

For an arithmetic progression, \(a_n=a+(n-1)d\). Therefore, \(a_{14}-a_6=(14-6)d=8d\). Using the given values, \(94-38=56=8d\), so \(d=7\). If \(d=6\), the difference across 8 positions would be only \(48\), not the required \(56\). Exam tip: when two terms are given, divide the difference of their values by the difference of their term numbers.

In an AP, the difference between two terms equals the common difference times the gap in their term numbers. Here the terms are 6 positions apart and 49−25=24, so d=24/6=4. Exam tip: always subtract the term indices as well as the term values.

Consecutive AP terms have equal differences, so \(q-p=r-q\). Rearranging gives \(2q=p+r\), meaning the middle term is the average of the outer terms. Option B does not express this. Exam tip: test the middle term first.

Use a_n = a_1 + (n - 1)d. The two conditions become a_1 + 3d = 20 and a_1 + 10d = 69. Subtracting gives 7d = 49, so d = 7. Substitute this into the first equation: a_1 + 3(7) = 20, hence a_1 + 21 = 20 and a_1 = -1. Therefore option A is correct. Verification is straightforward: the fourth term is -1 + 3×7 = 20, while the eleventh term is -1 + 10×7 = 69. The positive alternatives would produce a fourth term greater than 20 when the common difference is 7, so they cannot satisfy both given conditions.

Here, the first term is \(a=27\) and the common difference is \(d=35-27=8\). The nth term is \(a_n=a+(n-1)d\). Thus, \(123=27+(n-1)8\), so \((n-1)8=96\), \(n-1=12\), and \(n=13\). Option 12 is incorrect because it is the value of \(n-1\), not the term number \(n\). Exam tip: when finding the term number, remember to add 1 after calculating \(n-1\).

In option A, consecutive differences are \(2x-x=x\) and \(3x-2x=x\), so the common difference is fixed. Option D has changing differences. Exam tip: compare two successive differences to identify an AP.

Here, the first term is \(a=12\), the common difference is \(d=18-12=6\), and the number of terms is \(n=25\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), \(S_{25}=\frac{25}{2}[2(12)+24\times6]=\frac{25}{2}(168)=2100\). Therefore, option C is correct. Taking \(24\) as the first term would be incorrect; it is the value of \((n-1)\). Exam tip: always check \(n-1\) carefully in the sum formula.