Arithmetic Progression

From (a_2) to (a_8), the increase over (6) gaps is (30), so (d=5), hence (a_5=9+3(5)=24). Find the common difference first.

The common difference is (4), so after (14) the next terms are (18) and (22). Keep adding the common difference for next terms.

There are (4) gaps between (a_6) and (a_2), so the difference is (4\times11=44). For term difference look at the difference of positions.

For this arithmetic progression, the first term is \(a=27\) and the common difference is \(d=30-27=3\). Using \(a_n=a+(n-1)d\), we get \(a_n=27+(n-1)\times3=3n+24\). Hence, option A is correct. In option C, putting \(n=1\) gives 30, not the first term 27. Exam tip: always test a general term with \(n=1\) to verify the first term.

(a_7=2) and (a_8=-3), so the first negative term is the (8)th. Check carefully for the first term that goes below zero.

The common difference of an arithmetic progression is the difference between consecutive terms. Here, \(9-4=5\) and \(14-9=5\), so the common difference is 5. The number 4 is the first term, not the difference. Exam tip: subtract the first term from the second term to find the common difference.

(7) is added each time, so the next term is (32+7=39). In exams, identify the common difference first.

In an arithmetic progression, the second term is found by adding the common difference to the first term. Thus, second term = 9 + 4 = 13. Option 12 would require adding an incorrect difference, so it is not correct. Exam tip: To find the next term of an AP, add the common difference to the preceding term.

Each term decreases by (5), so the common difference is (-5). In exams, write the difference as negative for a decreasing progression.

A sequence is an arithmetic progression when the same number is added to obtain each next term. That fixed number is called the common difference. It is enough to compare consecutive terms, but checking all displayed pairs makes the pattern clear and prevents confusing the first term with the difference.

Here, \(13-8=5\), \(18-13=5\), and \(23-18=5\). Every difference is equal, so the sequence is an arithmetic progression with common difference 5. Hence choice A is correct. Choice B gives the first term rather than the change between terms. Choices C and D are incorrect because the differences are equal and positive, showing that the terms increase regularly.

In an arithmetic progression, the difference between every pair of consecutive terms must be the same. Here, the differences are \(4-2=2\), \(7-4=3\), and \(11-7=4\). Since these differences are unequal, the sequence is not an arithmetic progression. Increasing or positive terms alone do not make a sequence an AP. Exam tip: Find two or three consecutive differences to check whether a sequence is an AP.

An arithmetic progression is a sequence in which the difference between consecutive terms remains constant. However, to identify its first term, we do not need to calculate that difference. The terms are written in their order, and the first number displayed is the first term. In the sequence 16, 22, 28, 34, the number 16 appears first, so the first term is 16. Therefore, choice B is correct.

The common difference can be checked as \(22-16=6\), \(28-22=6\), and \(34-28=6\), confirming that the sequence is an arithmetic progression. This difference does not change the identity of the first term. The numbers 22 and 28 are later terms, while 6 is only the common difference, not a term in the displayed progression.

The general term of an arithmetic progression is


a_n=a+(n-1)d
a

. For the third term,


a_3=7+(3-1) imes5=7+10=17
a

. Therefore, 17 is correct. The value 15 is only the second term because the common difference 5 is added once. Exam tip: for the third term, add the common difference twice to the first term.

The common difference is (-4), so the next term is (30-4=26). In exams, apply the negative difference in decreasing terms.

The common difference is (6), so the fifth term is (31+6=37). In exams, extend the terms in order.

The difference between (21) and (28) is (7), so the missing term is (21-7=14). In exams, apply the difference backward too.

This is a sequence of multiples of (8), so (a_n=8n). In exams, identify multiple-based progressions as (kn).

Given \(a_n=5n+1\), substitute \(n=6\) for the sixth term: \(a_6=5\times 6+1=30+1=31\). Therefore, 31 is correct. The value 29 could result from subtracting 1 instead of adding it, so it is not correct. Exam tip: For a term \(a_n\), first substitute the value of \(n\), then perform multiplication and addition in order.

Here, the first term is \(a=9\) and the common difference is \(d=13-9=4\). Thus, \(a_n=a+(n-1)d=9+4(n-1)=4n+5\). Therefore, option B is correct. In option C, putting \(n=1\) gives the first term as \(-1\), not 9. Exam tip: identify the first term and common difference, then apply \(a_n=a+(n-1)d\) directly.

In an arithmetic progression, the second term equals the first term plus the common difference. Thus, the second term is 22 + (-6) = 16. Getting 18 would mean subtracting 4 from 22, but the given difference is 6. Exam tip: a negative common difference means that successive terms decrease.