Arithmetic Progression

An arithmetic progression is formed by repeatedly adding the same common difference. The first term is \(a=9\), and the common difference is \(d=0\). Therefore the second term is \(9+0=9\), and the third term is again \(9+0=9\). In fact, adding zero never changes a number, so every term of this progression remains 9.

The first option alternates incorrectly, while the second and fourth options show sequences with nonzero changes. The defining feature here is not merely the starting value 9 but the instruction that the difference is zero. Hence the first three terms are \(9,9,9\), which is exactly option C. The supplied answer and the calculation agree.

The difference between (10) and (14) is (4), so the missing term is (10-4=6). In exams, apply the common difference backward too.

In the linear rule (a_n=5n-2), the coefficient of (n) is the difference (5). In exams, take (k) as (d) in (kn+c).

Here, the first term is \(a=12\) and the common difference is \(d=15-12=3\). The \(n\)th term of an arithmetic progression is \(a_n=a+(n-1)d\). Therefore, \(a_6=12+(6-1)\times3=27\). Option 30 results from incorrectly using \(n\) instead of \(n-1\). Exam tip: Since the first term is \(a_1\), the formula contains \(n-1\).

In option A, consecutive differences are \(5-2=3\), \(8-5=3\), and \(11-8=3\), so it is an AP. In option D, the differences change. Exam tip: always check consecutive differences.

In the sequence (10, 8, 6, 4), 2 is subtracted to get each next term: 8 - 10 = -2, 6 - 8 = -2, and 4 - 6 = -2. Hence, its common difference is -2. Options A and C have positive common differences, while option D has common difference 0. Exam tip: Find the common difference by calculating next term − previous term.

Given \(a_n=11-2n\). Substituting \(n=3\), we get \(a_3=11-2(3)=11-6=5\). Therefore, option B is correct. The value \(7\) would result from subtracting only \(n\) instead of \(2n\), which does not follow the given rule. Exam tip: Substitute the value of \(n\) first, then perform multiplication and subtraction.

At (n=1) it gives (9), and at (n=2) it gives (7), so (a_n=11-2n). In exams, choose the rule after checking the first term.

The middle term is the average of (5) and (15), so (x=\frac{5+15}{2}=10). In exams, take the average for the middle term in three-term progressions.

In this arithmetic progression, the first term is \(a=18\) and the common difference is \(d=24-18=6\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_7=18+(7-1)\times6=18+36=54\). Hence, 54 is correct. The value 48 is the sixth term because the common difference is added only five times. Exam tip: for the \(n\)th term, use \((n-1)\), not \(n\).

An arithmetic progression is a sequence in which the same number is added to obtain each next term. The first term is the term at the beginning, and the common difference is found by subtracting one term from the next. In option A, the sequence starts with 6 and continues as 10, 14, and 18. Each step adds 4, so it has first term 6 and common difference 4.

The other choices fail at least one condition. Option B starts with 4, not 6, although its difference is 4. Option C starts with 6, but its difference is 6 because 12-6=6. Option D has a difference of 4 but starts with 10. Thus only option A satisfies both requirements simultaneously. A quick check is \(10-6=4\), \(14-10=4\), and \(18-14=4\).

The common difference of an arithmetic progression is the fixed amount by which one term changes to become the next term. It is calculated by subtracting a term from the following term. In this sequence, the first two terms are 5 and 9, so the difference is \(9-5=4\). Checking the next pairs gives \(13-9=4\) and \(17-13=4\), confirming that the difference remains constant.

Therefore the common difference is 4, which is option B. The number 3 in option A is not the change between consecutive terms, and 5 is merely the first term, not the difference. The number 9 is the second term. A reliable method is to subtract the first term from the second and then check one more pair if necessary, especially when identifying an arithmetic progression.

The nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Here, \(a=7\), \(d=3\), and \(n=5\), so \(a_5=7+(5-1)\times3=7+12=19\). Therefore, 19 is correct. Choosing 18 would mean adding 3 only three times, but there are four common differences from the first term to the fifth term. Exam tip: always use \(n-1\) when finding the nth term.

Each time (4) is subtracted, so the common difference is (-4). A decreasing sequence can also be an arithmetic progression.

The nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Therefore, \(a_4=11+(4-1)(-2)=11-6=5\). Hence, 5 is correct. The answer 7 results from treating the negative common difference incorrectly. Exam tip: always retain the negative sign of \(d\) while substituting in the formula.

(a_n=6+(n-1)4), and (6+4(n-1)=38) gives (n=9). When position is asked, set the given term equal to (a_n).

The governing definition is that an arithmetic progression has the same difference between every pair of consecutive terms. For option B, 5 − 1 = 4, 9 − 5 = 4, and 13 − 9 = 4, so the common difference is constant and the sequence is an arithmetic progression. Therefore option B is correct. Option A doubles each term, showing a constant ratio rather than a constant difference. Option C also doubles each term, so it is geometric, not arithmetic. In option D, the terms are consecutive squares: 1, 4, 9, 16; its differences are 3, 5, and 7, which are not equal. Checking differences, rather than merely observing that the terms increase, is the reliable test.

For an arithmetic progression, \(a_n=a_1+(n-1)d\). Thus, \(a_6=4+5d\). Substituting the given values gives \(24=4+5d\), so \(5d=20\) and \(d=4\). Option 5 is incorrect because there are 5 gaps between the first and sixth terms, not 4. Exam tip: always use \((n-1)d\) in the formula for \(a_n\).

In the linear form (a_n=2n+5), the coefficient of (n) is (2). In an arithmetic progression, this coefficient is the common difference.

Here (a=15) and (d=-3), so (a_7=15+6(-3)=-3). In a decreasing progression, terms may go below zero.