Arithmetic Progression

In an arithmetic progression, \(a_n=a_1+(n-1)d\). Hence, \(a_7=a_1+6d\). Substituting the given values, \(52=a_1+6\times7=a_1+42\), so \(a_1=10\). If 12 were used, the seventh term would be \(12+42=54\), not 52. Exam tip: the number of common differences before the \(n\)th term is always \(n-1\).

The position difference is (10) and (d=8), so the difference is (10\times8=80). In exams, use ((m-n)d) directly for the difference of two terms.

The first eight terms are (7,17,27,37,47,57,67,77), and their sum is (336). In exams, you can verify the sum by listing terms from the rule.

Here, the first term is \(a=23\) and the common difference is \(d=31-23=8\). The \(n\)th term is \(a+(n-1)d\). Thus, \(23+(n-1)\times 8=111\) gives \((n-1)=11\), so \(n=12\). Therefore, 111 is the 12th term. The 11th term is \(23+10\times8=103\), so it is not correct. Exam tip: identify \(a\) and \(d\) first, then equate the given value to \(a_n\).

For an arithmetic progression, \(a_n=a+(n-1)d\). Thus, \(a_2=a+d=16\) and \(a_9=a+8d=65\). Subtracting the two equations gives \(7d=49\), so \(d=7\). Substituting this into \(a+d=16\) gives \(a=9\). Therefore, the correct answer is 9. Option 8 can result from an incorrect substitution after finding the common difference. Exam tip: form equations for the given terms and subtract them first to find \(d\) quickly.

Here, the first term is \(a=64\) and the common difference is \(d=56-64=-8\). The \(n\)th term is \(a_n=a+(n-1)d\). For the zero term, \(64+(n-1)(-8)=0\), so \(n-1=8\) and hence \(n=9\). Therefore, the ninth term is zero. The eighth term is \(8\), not zero. Exam tip: To find the position of a specified term, write the \(a_n\) formula first and substitute the given value.

The formula for an arithmetic progression is \(a_n=a+(n-1)d\). Hence, \(a_{12}=a+11d\). Substituting the given values, \(88=11+11d\), so \(11d=77\) and \(d=7\). Therefore, option C is correct. If \(d=8\), the twelfth term would be \(99\), not the given value. Exam tip: the coefficient of \(d\) in the \(n\)th term is always \(n-1\).

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.

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.

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.