nth term

For an AP, \(a_n=a+(n-1)d\). Substituting \(a=7\) and \(d=4\) gives option A. In option B, putting \(n=1\) gives 11 rather than the first term 7. Exam tip: always test the rule at \(n=1\).

This is an arithmetic progression with first term \(a=15\) and common difference \(d=15\). Thus, \(a_n=a+(n-1)d=15+(n-1)15=15n\). In option B, putting \(n=1\) gives the first term as 0, so it is incorrect. Exam tip: Substitute \(n=1\) in a proposed rule to check whether it gives the first term.

In an AP, each term is obtained by adding the fixed common difference \(d\) to the previous term. Thus, \(a_n=a_{n-1}+d\), so \(d=a_n-a_{n-1}\). A sum or ratio does not give the common difference. Exam tip: subtract consecutive terms.

The governing concept is the nth term of an arithmetic progression. Consecutive terms differ by a constant amount: 9 − 6 = 3, 12 − 9 = 3, and 15 − 12 = 3. Therefore, the first term is a₁ = 6 and the common difference is d = 3. Using aₙ = a₁ + (n − 1)d, we obtain aₙ = 6 + (n − 1)3 = 6 + 3n − 3 = 3n + 3. Hence option A is correct. Substitution confirms the rule: n = 1 gives 6, n = 2 gives 9, n = 3 gives 12, and n = 4 gives 15. Option B begins with 3, option C begins with 0, and option D begins with 7 and has difference 1, so each distractor fails either the first-term test or the common-difference test.

To find the fourth term, substitute \(n=4\): \(a_4=\frac{3\times4}{2}=\frac{12}{2}=6\). Hence, the correct answer is 6. Option 8 would result from not dividing \(3\times4\) by 2. Exam tip: substitute the term number carefully in the formula for the \(n\)th term, then simplify.

Direct answer: Option A, \\(a_n=\\frac{3n}{2}\\), is correct. Look at the terms: \\(\\frac32,3,\\frac92,6\\). They can be written as \\(\\frac32\cdot1,\\frac32\cdot2,\\frac32\cdot3,\\frac32\cdot4\\). Therefore the nth term is \\(\\frac32\cdot n=\\frac{3n}{2}\\). Check: at \\(n=1\\), it gives \\(\\frac32\\); at \\(n=2\\), it gives 3; at \\(n=4\\), it gives 6. Option A matches all terms. Option B, \\(\\frac n3\\), gives \\(\\frac13\\) as the first term and is too small. Option C, \\(3n\\), gives 3 as the first term and is twice the required value. Option D, \\(n+\\frac32\\), gives \\(\\frac52\\) as the second term, not 3. Memory cue: when terms are equal to a fixed fraction times 1, 2, 3, use that multiplier times n.

For the third term, substitute n=3: a_3=2^3+1=8+1=9. Therefore, 9 is correct. Option 8 is only the value of 2^3; the +1 in the formula still has to be added. Exam tip: when finding an nth term, substitute the value of n first, then evaluate powers and other operations in order.

The governing idea is the general term of an arithmetic progression. The difference between successive terms is constant: 9 − 5 = 4, 13 − 9 = 4, and 17 − 13 = 4. Thus a₁ = 5 and d = 4. Applying aₙ = a₁ + (n − 1)d gives aₙ = 5 + (n − 1)4 = 5 + 4n − 4 = 4n + 1. Therefore, option A is correct. A direct check gives a₁ = 4(1) + 1 = 5, a₂ = 9, a₃ = 13, and a₄ = 17, exactly as required. Option B starts at 4, option C starts at 3, and option D starts at 5 but increases by 1 instead of 4. Consequently, only option A satisfies both the first-term condition and the common-difference condition.

The nth term of an AP is \(a_n=a+(n-1)d\). Here, \(a=7\) and \(d=3\), so option A is correct. In option B, the first term becomes 10. Exam tip: identify \(a\) and \(d\) before choosing the formula.

This is an arithmetic progression with first term 20 and common difference 20. Therefore, \(a_7=20+(7-1)\times20=140\). The number 120 is the sixth term, so it is a close but incorrect option. Exam tip: write the term number or use \(a_n=a+(n-1)d\) to avoid counting errors.

This is an arithmetic progression because \(a_{n+1}-a_n=[5(n+1)+2]-(5n+2)=5\), a constant difference. A geometric progression requires a constant ratio instead. Exam tip: a form \(a_n=dn+c\) indicates an AP.

This question uses the nth-term formula for an arithmetic progression. The consecutive differences are constant: 10 − 4 = 6, 16 − 10 = 6, and 22 − 16 = 6. Hence the first term is a₁ = 4 and the common difference is d = 6. Substituting into aₙ = a₁ + (n − 1)d gives aₙ = 4 + (n − 1)6 = 4 + 6n − 6 = 6n − 2. Therefore, option A is correct. Checking n = 1, 2, 3, and 4 produces 4, 10, 16, and 22 respectively. Option B begins at 8, option C begins at 10, and option D has common difference 2. Since these alternatives fail the first term, the difference, or both, they cannot represent the given sequence.

To find the ninth term, substitute \(n=9\) in the rule: \(a_9=\frac{9}{3}=3\). Therefore, the correct answer is 3. Option 6 could result from incorrectly treating the operation as \(9-3\) instead of \(\frac{9}{3}\). Exam tip: first substitute the required value of \(n\) into the nth-term formula, then simplify.

Each term is obtained by dividing (n) by (3), so (a_n=\frac{n}{3}). In exams, view fractions in the same form.

For the second term, substitute n=2. Thus, a_2=5^2=25, so 25 is correct. The value 125 equals 5^3, so it is the third term. Exam tip: carefully substitute the required term number for n in the nth-term formula.

The formula gives the value of a term when its position number is known. For the fourth term, the position is represented by putting \\(n=4\\) into the given rule. The expression has a product in the numerator and division by 2, so both the multiplication and the division must be completed carefully. This type of formula is useful because it finds a term directly without listing all earlier terms.

Substitute 4: \\(a_4=\\frac{4(4+1)}{2}=\\frac{4\\times5}{2}=\\frac{20}{2}=10\\). Therefore, the fourth term is 10, so option D is correct. The value 6 would result from using an unsuitable expression, while 8 or 9 do not follow from the stated formula. The key step is to use the position number 4, not the term value from some other pattern.

Its rule is (a_n=n(n+1)), and (6\times7=42). In exams, match the given term with the rule to find the term number.

To find the fourth term, substitute \(n=4\) in the formula: \(a_4=4+8=12\). Therefore, the correct answer is 12. Option 13 would result from using \(n=5\), so it is the fifth term. Exam tip: the subscript \(n\) in \(a_n\) indicates the term number.

The first term is 7 and the common difference is 4. Thus \(a_n=7+(n-1)\times4=4n+3\). Option B gives 1 when \(n=1\), not 7. Exam tip: verify both the first term and common difference.

For \(n^2+1\), the difference of consecutive terms is \((n+1)^2+1-(n^2+1)=2n+1\), which changes with n. An AP needs a constant difference. Exam tip: any form \(pn+q\) represents an AP.

In a square-number sequence, the nth term is \(n^2\). Therefore, the eighth term is \(8^2=64\). \(49=7^2\) is the seventh term, while \(81=9^2\) is the ninth term. Exam tip: square the term number to find a term in this sequence.

For the fifth term, substitute n = 5. Thus, a₅ = 5² + 3 = 25 + 3 = 28, so 28 is the correct option. The value 27 would result from using 5² + 2, which is not the given rule. Exam tip: while finding an nth term, first substitute the correct value of n and then evaluate the power.

The governing concept is the nth-term formula for an arithmetic progression. The first term is 2, and each successive term increases by 4: 6 − 2 = 4, 10 − 6 = 4, and 14 − 10 = 4. Thus a₁ = 2 and d = 4. Using aₙ = a₁ + (n − 1)d, we get aₙ = 2 + (n − 1)4 = 2 + 4n − 4 = 4n − 2. Therefore, option A is correct. Substitution confirms that n = 1 gives 2, n = 2 gives 6, n = 3 gives 10, and n = 4 gives 14. Option B starts at 6, while options C and D do not produce the required first term and common difference, so none of them can describe the sequence.

The governing concept is finding the position of a specified value in an arithmetic sequence. The first term is 8 and the common difference is also 8, so the nth term is aₙ = 8 + (n − 1)8 = 8n. To locate 72, set 8n = 72 and divide both sides by 8: n = 9. Therefore, 72 is the ninth term, so option C is correct. Listing the sequence confirms this: 8, 16, 24, 32, 40, 48, 56, 64, 72. The seventh term is 56, the eighth is 64, and the tenth is 80. Hence options A, B, and D correspond to different values and cannot be correct.

Given \(a_n=2^n\), substitute \(n=6\) for the sixth term: \(a_6=2^6=64\). Therefore, 64 is correct. \(2^5=32\) is the fifth term, so it is a close but incorrect option. Exam tip: To find a term, substitute its position directly for \(n\) in the nth-term formula.