Arithmetic Progression

For an arithmetic progression, \(a_n=a_1+(n-1)d\). Thus, \(a_6=17+(6-1)d\), so \(47=17+5d\). Hence, \(5d=30\) and \(d=6\). There are 5 gaps between the first and sixth terms, not 6. Exam tip: always use \((n-1)d\) in the formula for \(a_n\).

The governing concept is the nth-term form of an arithmetic progression. For an AP, a_n=a+(n-1)d, so the coefficient of n in a linear nth-term rule is the common difference d. Here a_n=55-5n, and the coefficient of n is -5; therefore d=-5. This can also be checked directly: a_1=55-5=50 and a_2=55-10=45, so a_2-a_1=45-50=-5. Hence option A is correct. Option B ignores the negative sign, while 50 is the first term rather than the difference and 55 is the constant appearing before the n-term, not d.

For an arithmetic progression, the fourth term is \(a_4=a+3d\). Here, \(36=21+3d\), so \(3d=15\) and \(d=5\). Therefore, the correct answer is 5. Choosing 3 would be incorrect: there are three gaps from the first to the fourth term, and dividing 15 by 3 gives 5. Exam tip: use \(a_n=a+(n-1)d\) and carefully count the number of gaps between terms.

The first term is \(a=4\), and the common difference is \(d=13-4=9\). The \(n\)th term of an arithmetic progression is \(a_n=a+(n-1)d\). Therefore, \(a_{10}=4+(10-1)\times9=4+81=85\). Hence, option C is correct. The value \(76\) is \(4+8\times9\), which is the 9th term, not the 10th term. Exam tip: use \(n-1\) differences to find the \(n\)th term.

In an arithmetic progression, the second term is \(a_2=a_1+d\). Therefore, \(a_1=a_2-d=18-6=12\). Hence, the correct answer is \(12\). Getting \(24\) would mean adding the common difference instead of subtracting it, which is incorrect here. Exam tip: subtract the common difference when moving back to the first term.

Here (a=60) and (d=-6), so (a_{12}=60+11(-6)=-6). In a decreasing progression, terms can become negative.

In option C, the first term is 12. The differences between consecutive terms are 8-12=-4, 4-8=-4, and 0-4=-4, so it is an AP with common difference -4. Option D also has common difference -4, but its first term is 16. Exam tip: To identify an AP, check the first term and at least two consecutive differences.

To find the first term, substitute
n=1
. Thus,
a_1=6(1)-1=5
. Therefore, the first term is 5. Option 6 is only the coefficient in the formula, not the first term. Exam tip: For any sequence, put
n=1
in the general-term formula to get its first term.

The first four terms are 12, 20, 28, and 36. Therefore, their sum is \(12+20+28+36=96\). Option 92 is the sum of only the first three terms, \(12+20+28\), so it is incorrect. Exam tip: When an arithmetic progression has only a few terms, adding them directly is often quickest.

The nth-term formula of an arithmetic progression is \(a_n=a+(n-1)d\). Thus, \(a_6=34+(6-1)(-7)=34-35=-1\). Therefore, \(-1\) is correct. The value \(6\) results from incorrectly using 4 instead of \(n-1=5\) for the sixth term. Exam tip: when \(d\) is negative, check the sign after multiplication carefully.

From the fourth to the eighth term there are (4) gaps and the increase is (24), so (d=6). Count the number of gaps correctly.

Given \(a_n=8n+5\), \(a_2=8\times2+5=21\) and \(a_5=8\times5+5=45\). Therefore, \(a_2+a_5=21+45=66\). A value such as \(62\) can result from evaluating one of the terms incorrectly. Exam tip: substitute the value of \(n\) in each term separately before adding them.

The governing concept is the relationship between a specified term and the first term of an arithmetic progression: aₙ = a + (n − 1)d. For the sixth term, a₆ = a + 5d. The given values are a₆ = 41 and d = 5, so 41 = a + 5×5 = a + 25. Subtracting 25 from both sides gives a = 16. Therefore, option B is correct. A quick check is the sequence 16, 21, 26, 31, 36, 41, whose sixth term is indeed 41. Options A, C, and D result from subtracting the wrong multiple of the common difference.

The common difference of an arithmetic progression is the difference between two consecutive terms. Here, both the second and first terms are 12, so \(d=12-12=0\). Therefore, the correct answer is 0. The number 12 is the value of each term, not the common difference. Exam tip: use \(d=a_2-a_1\) to find the common difference.

The first term is (22) and (d=7), so (a_n=22+(n-1)7=7n+15). Remember (n-1) while forming the general term.

From (a_5) to (a_{10}), there are (5) gaps, so (a_{10}=33+5(4)=53). Moving forward from a nearby term is a quick method.

The first term is (20), and subtracting (5) each time gives (20,15,10,5). Form terms directly from the given (a) and (d).

In an arithmetic progression, (2(2x)=(x+1)+(3x-5)), which gives (x=6). The middle term is the average of the two surrounding terms.

(a_8=3) and (a_9=-2), so the first negative term is the (9)th. Check the first term that goes below zero.

For the fourth term, substitute n=4 in the given rule: \(a_4=17-3(4)=17-12=5\). Therefore, the correct answer is 5. The value 8 is obtained when \(n=3\), so it is the third term, not the fourth. Exam tip: In an \(a_n\) question, substitute the required term number directly for n.