Arithmetic Progression

In an arithmetic progression, d is the common difference between consecutive terms. For option A, every term is 4, so 4 − 4 = 0 throughout and d = 0. Option B has d = 1, option C has d = 5, and option D has d = −3. A constant sequence is therefore a valid arithmetic progression with zero common difference, making option A correct.

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.

In this progression (d=0), so every term is (4). In a constant arithmetic progression, all terms are equal.

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 first term is (21) and (d=7), so (a_n=21+(n-1)7=7n+14). You can also check an option by putting (n=1).

This is an AP with first term 18 and common difference 2. The 6th term is 18 + (6 - 1) × 2 = 28. Choosing 30 gives the 7th-row value instead. Exam tip: always use n - 1 in the nth-term formula.

(5) is added each time, so the next term is (27+5=32). In exams, identify the common difference first.

In an arithmetic progression, each next term is obtained by adding the common difference to the previous term. Therefore, the second term is 6 + 7 = 13. The value 14 would result from adding 7 twice. Exam tip: remember that the second term is always a + d.

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

The defining concept of an arithmetic progression is that the difference between every pair of consecutive terms remains constant. Check the sequence directly: 12 − 6 = 6, 18 − 12 = 6, and 24 − 18 = 6. Since all consecutive differences are equal, the sequence is an arithmetic progression with common difference 6. Therefore option A is correct. Option B confuses the second term with the common difference. Option C is false because the calculated differences are equal, and option D is false because the terms are increasing, not decreasing. A sequence may have a positive, negative, or zero common difference; here it is positive.

For the given sequence, the consecutive differences are \(5-2=3\), \(10-5=5\), and \(17-10=7\). Since these differences are not equal, it is not an arithmetic progression. Merely having increasing terms does not make a sequence an arithmetic progression. Exam tip: always check the difference between consecutive terms to identify an AP.