Arithmetic Progression

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 governing test for an arithmetic progression is equality of consecutive differences. For the sequence shown, the first difference is 5 − 2 = 3, the second is 9 − 5 = 4, and the third is 14 − 9 = 5. Since 3, 4, and 5 are not equal, one constant common difference does not exist; therefore option A is correct. An arithmetic progression may begin with any number, including 2, so option B gives no valid reason. Its terms may increase, but increasing terms alone do not make a sequence arithmetic, so C is false. The number of displayed terms is irrelevant, making D false. The changing differences are the decisive evidence.

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 \(a_1\).

For an arithmetic progression, \(a_n=a_1+(n-1)d\). Thus, \(32=12+(6-1)d=12+5d\). Hence \(5d=20\), so \(d=4\). There are 5 gaps, not 6, between \(a_1\) and \(a_6\). Exam tip: while finding the common difference from two terms, use the difference between their term numbers.

The governing concept is the sum of a finite arithmetic progression. The first term is a = 5 and the common difference is d = 12 − 5 = 7. The fifth term is a5 = a + 4d = 5 + 28 = 33. Applying the sum formula, S5 = 5/2 [2(5) + (5 − 1)(7)] = 5/2 (10 + 28) = 5/2 × 38 = 95. Directly listing the terms gives 5, 12, 19, 26, and 33; their total is also 95. Thus option C is correct. The other options result from an arithmetic or formula error, such as omitting the fifth term or using an incorrect difference.

The governing concept is the general form of an arithmetic progression. If a_n = 42 − 6n, then increasing n by 1 changes the term by −6: a_(n+1) − a_n = [42 − 6(n+1)] − [42 − 6n] = 42 − 6n − 6 − 42 + 6n = −6. Therefore the common difference is −6, so option A is correct. The negative sign shows that the progression decreases by 6 at each step. Option B ignores that sign, while 36 and 42 are merely numbers appearing through multiplication or the constant term and are not the difference between consecutive terms. Checking values also confirms it: a1 = 36 and a2 = 30, so a2 − a1 = −6.

An arithmetic progression is identified by a constant difference between consecutive terms. In option C, 11 − 7 = 4, 15 − 11 = 4, and 19 − 15 = 4, so every consecutive difference is equal and the sequence is an arithmetic progression. Therefore option C is correct. Option A doubles each term, giving changing differences 3, 6, and 12; it is geometric rather than arithmetic. Option B consists of successive square numbers, whose differences are 5, 7, and 9, so they are not constant. Option D also doubles each term and has differences 2, 4, and 8. Equal ratios or a recognisable pattern are not enough; the defining test here is equal subtraction results.

In an arithmetic progression, the fourth term is reached after three equal gaps from the first term. Thus, \(a_4=a_1+3d\). Here, \(30=18+3d\), so \(3d=12\) and \(d=4\). If the common difference were 3, the fourth term would be \(18+3\times3=27\), not 30. Exam tip: from the first term to the \(n\)th term, there are \(n-1\) common differences.

The first term is \(a=11\) and the common difference is \(d=17-11=6\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{12}=11+(12-1)\times6=11+66=77\). Getting 73 usually results from using an incorrect number of common differences. Exam tip: from the first term to the \(n\)th term, the common difference is added \(n-1\) times.

The \(n\)th term of an arithmetic progression is \(a_n=a+(n-1)d\). Hence, \(a_{10}=7+(10-1)\times4=7+36=43\). Although 43 is correct, there are 9 intervals from the first term to the 10th term, so 4 is added 9 times, not 10 times. Exam tip: use \((n-1)\) when finding the \(n\)th term.

In option C, the first term is 6. The differences between consecutive terms are 3 - 6 = -3, 0 - 3 = -3, and -3 - 0 = -3, so its common difference is -3. Option D has common difference -3, but its first term is 9. Exam tip: To find the common difference of an AP, subtract the first term from the second term.

To find the first term, put n=1. Thus, a₁=7(1)-2=5. Therefore, the correct answer is 5. The number 7 is the coefficient of n, not the first term. Exam tip: To find the first term of a sequence, substitute n=1.

Here, the first term is \(a=16\) and the common difference is \(d=5\). The \(n\)th term is \(a_n=a+(n-1)d\). So, \(16+(n-1)\times5=56\) gives \(n-1=8\), hence \(n=9\). Therefore, 56 is the 9th term. The close distractor, the 8th term, is incorrect because the 8th term is \(16+7\times5=51\). Exam tip: To find the position of a given term in an AP, start with \(a_n=a+(n-1)d\).

The first four terms of the arithmetic progression are 9, 15, 21, and 27. Their sum is \(9+15+21+27=72\). Therefore, 72 is the correct answer. A value such as 78 can result from an incorrect addition. Exam tip: For a small number of terms, direct addition is quickest; for more terms, use \(S_n=\frac{n}{2}[2a+(n-1)d]\).

The nth-term formula for an arithmetic progression is \(a_n=a+(n-1)d\). Therefore, \(a_7=25+(7-1)(-5)=25-30=-5\). Hence, \(-5\) is correct. \(0\) would result from subtracting the common difference only 5 times instead of 6 times. Exam tip: when \(d\) is negative, each successive term decreases.

From the third to the sixth term there are (3) gaps and the increase is (15), so (d=5). Count the number of gaps correctly.

Given \(a_n=5n+6\), \(a_3=5(3)+6=21\) and \(a_5=5(5)+6=31\). Therefore, \(a_3+a_5=21+31=52\). An option such as 48 may result from incorrectly handling the \(+6\) term. Exam tip: substitute the value of \(n\) separately to find each required term before adding.