Arithmetic Progression

The average of the first three terms is (\frac{10+18+26}{3}=18). In an arithmetic progression, the average of three consecutive terms is the middle term.

This is an AP with \(a=18\), \(d=3\), and \(n=12\). Thus, \(a_{12}=18+(12-1)\times3=51\). Getting 54 means adding the difference 12 times instead of 11. Exam tip: always use \(n-1\) gaps.

The first term is \(a=5\), and the common difference is \(d=14-5=9\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{14}=5+(14-1)\times9=5+117=122\). Hence, option D is correct. Option C, \(119\), results from using too small an increase instead of the required \(13d\) for the 14th term. Exam tip: for the \(n\)th term, remember that there are \(n-1\) common differences after the first term.

In (8,16,24,32,\ldots), (8) is added each time. To find the common difference, check the difference of two consecutive terms.

The first (5) terms are (44,38,32,26,20), and their sum is (160). In a decreasing progression, also list the terms in order and add.

From (a_7) to (a_4), move back (3) gaps, so (a_4=58-3(8)=34). When moving backward, subtract the common difference.

Here, the first term is 35 and the common difference is \(-7\). Therefore, \(a_n=35+(n-1)(-7)\). Putting \(a_n=0\), we get \(35-7(n-1)=0\), so \(n=6\). Hence, 0 is the sixth term. The fifth term is 7, so it is not correct. Exam tip: To find a term’s position, substitute its value in \(a_n=a+(n-1)d\).

This is an arithmetic progression with first term 10 and common difference 2. The 12th term is \(a+(n-1)d=10+11\times2=32\) cm. Choosing 34 cm gives the 13th term instead. Exam tip: use \(n-1\), not \(n\), in the formula.

There are (6) gaps between (a_{12}) and (a_6), and (d=8), so the difference is (48). Term difference depends on the difference of positions.

The rule is \(a_n=16-2n\). Substituting \(n=1,2,3,4\) gives \(a_1=14\), \(a_2=12\), \(a_3=10\), and \(a_4=8\), respectively. Hence, the correct sequence is \(14,12,10,8\). Option B incorrectly treats \(16\) as the first term; however, at \(n=1\), the first term is \(14\). Exam tip: When a sequence is defined using \(a_n\), usually begin by substituting \(n=1\).

From (a_3) to (a_9), the increase over (6) gaps is (36), so (d=6), hence (a_6=14+3(6)=32). Find the common difference first.

The common difference is (7), so after (23) the next terms are (30) and (37). Keep adding the common difference for next terms.

There are (5) gaps between (a_7) and (a_2), so the difference is (5\times13=65). For term difference, look at the difference of positions.

The first term is \(a=31\), and the common difference is \(d=35-31=4\). Hence, \(a_n=a+(n-1)d=31+4(n-1)=4n+27\). Therefore, option A is correct. In option C, putting \(n=1\) gives 35, whereas the first term must be 31. Exam tip: Always substitute \(n=1\) in a proposed general term to verify the first term.

The increase over three gaps is (12), so (d=4), then (a_1=19-3(4)=7). From middle terms first find the common difference.

In (6,12,18,24,30,\ldots), the fifth term is (30) and the common difference is (6). Check each option up to the fifth term.

In an AP, the difference between consecutive terms is a constant d; for example, 3, 7, 11 has common difference 4. Doubling terms indicates a geometric pattern, not an AP. Exam tip: subtract consecutive terms to test an AP.

The governing concept is the nth-term formula for an arithmetic progression: a_n = a + (n - 1)d. In the sequence 42, 36, 30, 24, each term decreases by 6, so the first term is a = 42 and the common difference is d = -6. For the eighth term, substitute n = 8: a_8 = 42 + (8 - 1)(-6) = 42 + 7(-6) = 42 - 42 = 0. Therefore option D is correct. Option A is only the common difference, option B reverses its sign, and option C does not follow the repeated subtraction pattern.

To find which term is 91, put \(a_n=91\) in the general term: \(4n+3=91\). Thus, \(4n=88\), so \(n=22\). Therefore, 91 is the 22nd term. The 21st term is \(4(21)+3=87\), so it is not correct. Exam tip: To find the term number of a given value, equate \(a_n\) to that value and solve for \(n\).

For an arithmetic progression, \(a_n=a+(n-1)d\). Here, \(a=14\) and \(a_{15}=84\), so \(84=14+(15-1)d=14+14d\). Thus, \(70=14d\), giving \(d=5\). If the common difference were 4, the fifteenth term would be \(14+14\times4=70\), not 84. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.