Arithmetic Progression

From (a_4) to (a_9), there are (5) gaps, so (a_9=18+5(3)=33). Moving forward from a nearby term is a quick method.

To find the first term, substitute \(n=1\). Thus, \(a_1=9(1)+2=11\), so 11 is correct. Option 9 is only the coefficient of \(9n\); the constant term 2 must also be added. Exam tip: To obtain the first term from any \(a_n\), always put \(n=1\).

The direct answer is option A: \((5,7,9,11)\). In an arithmetic progression, the first term is \(a=5\), and the common difference is \(d=2\), meaning add 2 to obtain each next term. Thus the first term is 5, the second is \(5+2=7\), the third is \(7+2=9\), and the fourth is \(9+2=11\). Option A follows both given conditions exactly. Option B starts with 2, not 5, and its difference is 3. Option C starts with 5 but adds 5 each time, so its common difference is 5 rather than 2. Option D starts with 7, not the required first term, although its difference is 2. The formula \(a_n=a+(n-1)d\) also gives \(a_n=5+2(n-1)\), confirming the list. Memory cue: first write the given first term, then repeatedly add the common difference.

From the third to the fifth term there are (2) gaps and the increase is (8), so (d=4). Count the number of gaps between terms.

The direct answer is option A: \(a_n=26-4n\). The first term is 22, and the common difference is \(18-22=-4\). The standard AP formula is \(a_n=a_1+(n-1)d\). Substitution gives \(a_n=22+(n-1)(-4)=22-4n+4=26-4n\). Checking: at \(n=1\), the formula gives 22; at \(n=2\), it gives 18; at \(n=3\), it gives 14; and at \(n=4\), it gives 10. Option A therefore matches. Option B, \(22-4n\), gives 18 at \(n=1\), so it starts incorrectly. Option C, \(4n+18\), increases and gives 22, 26, 30, so it has the wrong direction. Option D, \(26+4n\), also increases and gives 30 first. The important point is that a decreasing AP has a negative common difference. Memory cue: simplify \(a_1+(n-1)d\) carefully; the extra \(+4\) changes 22 into 26.

The average of the first three terms is (\frac{2+8+14}{3}=8). In an arithmetic progression, the average of three consecutive terms is the middle term.

This is an AP with \(a=20\), \(d=3\), and \(n=10\). So, \(S_{10}=\frac{10}{2}[2(20)+9(3)]=5(67)=335\). Finding only the 10th term does not give the total seats. Exam tip: use \(S_n\) when a total is asked.

The first term is \(a=17\), and the common difference is \(d=23-17=6\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(a_6=17+(6-1)\times6=17+30=47\). Therefore, 47 is correct. Note that 41 is the fifth term, not the sixth. Exam tip: add the common difference \(n-1\) times to obtain the \(n\)th term.

Given \(a_n=5n+3\), \(a_2=5(2)+3=13\) and \(a_4=5(4)+3=23\). Therefore, \(a_2+a_4=13+23=36\). The option 34 can result from an error while evaluating one of the terms. In exams, find each required term first and then add them.

Here, the first term is \(a=50\) and the common difference is \(d=-5\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(0=50+(n-1)(-5)\), giving \(n-1=10\) and hence \(n=11\). The 10th term is \(5\), not zero. Exam tip: To find a term’s position, substitute the given term for \(a_n\) in \(a_n=a+(n-1)d\).

The formula for the nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Therefore, \(a_8=2+(8-1)\times9=2+63=65\). Hence, 65 is correct. \(63\) is only the value of \(7d\); the first term \(a=2\) must also be added. Exam tip: for the nth term, the number of common differences is always \(n-1\).

In an arithmetic progression, each term is found by adding the common difference to the previous term. The given progression starts at 3 and increases by 10, so its first five terms are 3, 13, 23, 33, and 43. To find their sum, add them directly: \(3+13+23+33+43=115\). Thus the correct choice is option C.

The same result follows from the arithmetic-series formula \(S_n=\frac{n}{2}(a+l)\), where n is the number of terms, a is the first term, and l is the last term. Here n=5, a=3, and l=43, so \(S_5=\frac{5}{2}(3+43)=\frac{5}{2}\times46=115\). The value 105 would omit some increase, while 110 and 125 do not equal the required sum.

For an arithmetic progression, the nth term is \(a_n=a_1+(n-1)d\). Thus, \(a_6=25+(6-1)(-5)=25-25=0\). Therefore, 0 is the correct answer. The option 5 may result from adding \(-5\) only four times instead of five times to reach the sixth term. Exam tip: always use \((n-1)\) common differences to find \(a_n\), not \(n\) differences.

Given \(a_n=3n+8\), putting \(n=1\) gives \(a_1=11\). Each successive term increases by 3, so the progression is \((11,14,17,20,\ldots)\). Option A also has common difference 3, but its first term is 8, so it is not correct. Exam tip: substitute \(n=1\) to find the first term, then check the common difference.

Here, the first term is \(a=6\) and the common difference is \(d=15-6=9\). The \(n\)th term is \(a_n=a+(n-1)d\). So, \(69=6+(n-1)\times9\), giving \(n-1=7\) and \(n=8\). Hence, 69 is the 8th term. The 9th term would be \(78\), so option C is not correct. Exam tip: use \(n-1\), not \(n\), in the formula for the nth term of an AP.

From (a_5) to (a_2), move back three gaps, so (a_2=28-3(4)=16). When moving backward, subtract the common difference.

In an arithmetic progression, (2(x+3)=(x-2)+(2x-1)), which gives (x=9). The middle term is the average of the two adjacent terms.

The direct answer is B: 4. In an arithmetic progression, the nth-term formula is a_n=a+(n-1)d . Here every displayed term is 4, so the common difference is d=4-4=0 and the first term is a=4 . For the twentieth term, a_{20}=4+(20-1)(0)=4+0=4 . Another simple way is to notice that adding zero changes nothing, so the sequence remains 4, 4, 4, 4 for every position. Option A, 0, is the common difference, not the twentieth term. Option B is correct because the twentieth entry is still 4. Option C, 20, confuses the term number with the term's value. Option D, 80, incorrectly multiplies 4 by 20; multiplication is not the rule for this constant progression. Exam cue: distinguish the position n from the value of the term, and calculate d before using the formula.

Given \(a_n=18-2n\), \(a_3=18-2(3)=12\) and \(a_6=18-2(6)=6\). Therefore, \(a_3+a_6=12+6=18\). A value such as 16 can result from substituting an incorrect value of \(n\) in one term. Exam tip: find each required term separately before adding them.

The direct answer is option A: \(a_n=7n+14\). The first term is 21 and the common difference is \(28-21=7\). Apply \(a_n=a_1+(n-1)d\): \(a_n=21+(n-1)7=21+7n-7=7n+14\). Checking gives 21 when \(n=1\), 28 when \(n=2\), 35 when \(n=3\), and 42 when \(n=4\). Thus option A is exact. Option B, \(14n+7\), gives 21 first but then 35, so its difference is 14 rather than 7. Option C, \(7n+21\), gives 28 at \(n=1\), so it starts one step late. Option D, \(21n+7\), has difference 21 and gives 28 first. Substituting \(n=1\) is a quick way to reject formulas, but also check the difference or a second term. Memory cue: first term plus \((n-1)\) times the common difference.

The nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Here, \(a=9\), \(d=4\), and \(n=8\), so \(a_8=9+(8-1)\times4=9+28=37\). Therefore, 37 is the correct answer. An answer such as 35 usually results from using the wrong number of differences. Exam tip: multiply the common difference by \(n-1\), not by \(n\).

The first term is (35) and the difference is (-4), so (a_n=35+(n-1)(-4)=39-4n). In a decreasing progression the common difference is negative.

The formula for the nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Thus, \(a_5=6+(5-1)\times7=6+28=34\). Hence, 34 is correct. The common difference is added only four times to reach the fifth term from the first term, so 35 does not follow the AP rule. Exam tip: for the nth term, use \(n-1\) common differences.

Here, the first term is \(a=8\) and the common difference is \(d=5\). The \(n\)th term is \(a_n=a+(n-1)d\). So, \(8+(n-1)\times5=48\) gives \(5(n-1)=40\), hence \(n=9\). Therefore, 48 is the 9th term. The 8th term is \(43\), so it is not correct. Exam tip: To find the position of a given term, equate it to \(a_n\) and solve for \(n\).

Given \(a_n=3n+2\), substitute \(n=1,2,3,4\). This gives \(a_1=5\), \(a_2=8\), \(a_3=11\), and \(a_4=14\). Hence, the first four terms are \((5,8,11,14)\). Option C starts with 2, which is obtained by using \(n=0\), but the first term is found using \(n=1\). Exam tip: when a sequence is defined by \(a_n\), begin listing terms from \(a_1\).