nth term

Twice the position number \(n\) is \(2n\). Adding 5 gives \(a_n=2n+5\). Option \(2n-5\) incorrectly subtracts 5. Exam tip: translate “more than” as addition before choosing the formula.

This is an arithmetic progression with first term \(a=12\) and common difference \(d=12\). Thus, \(a_n=a+(n-1)d=12+(n-1)\times12=12n\). The expression \(a_n=12n+12\) gives 24 as the first term, so it is incorrect. Exam tip: Substitute \(n=1\) to check whether the rule gives the first term.

For the fourth term, substitute n=4. Thus, a_4=4^2+4=16+4=20. Therefore, 20 is the correct answer. The value 16 is only 4^2; the additional +4 must also be included. Exam tip: when finding an nth term, substitute the term number carefully in the formula.

The rule for this sequence is \(a_n=n(n+1)=n^2+n\). Substituting \(n=1,2,3,4\) gives \(2,6,12,20\), respectively. With \(a_n=n^2\), the second term would be \(4\), not the given \(6\). Exam tip: verify a proposed nth-term rule using the first few terms.

\(n\) indicates the position of a term, so \(a_4\) is the fourth term of the sequence. The value of \(a_n\) comes from its rule; the common difference is a separate quantity. Exam tip: read the subscript as the term number.

Each term in this sequence is a multiple of 10, so its nth term is \(a_n=10n\). Putting \(10n=90\) gives \(n=9\). Therefore, 90 is the ninth term. The tenth term would be 100, so it is not correct. Exam tip: for this sequence, divide the given number by 10 to find its term number quickly.

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 terms form an arithmetic progression because the common difference is constant: 9 − 6 = 3, 12 − 9 = 3, and 15 − 12 = 3. Use the arithmetic-progression formula aₙ = a₁ + (n − 1)d. With first term a₁ = 6 and common difference d = 3, we obtain aₙ = 6 + (n − 1)3 = 6 + 3n − 3 = 3n + 3. Therefore, option A is correct. Checking n = 1 gives a₁ = 3 + 3 = 6; n = 2 gives 9; n = 3 gives 12; and n = 4 gives 15. Option B gives 3 as the first term, option C gives 0, and option D gives 7, so those alternatives fail the first-term 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.

The terms are (\frac{3}{2}) times (n), so (a_n=\frac{3n}{2}). In exams, identify the fractional multiplier.

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 sequence is an arithmetic progression with first term 5 and common difference 4, since 9 − 5 = 4, 13 − 9 = 4, and 17 − 13 = 4. Apply aₙ = a₁ + (n − 1)d: aₙ = 5 + (n − 1)4 = 5 + 4n − 4 = 4n + 1. Thus option A is the correct general rule. The first-term check is decisive: for n = 1, 4(1) + 1 = 5; for n = 2, the value is 9; for n = 3, it is 13. Option C starts at 3, option D starts at 5 but then increases by only 1, and option B starts at 4, so these do not generate the full sequence.

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 is an arithmetic progression because the difference remains 6: 10 − 4 = 6, 16 − 10 = 6, and 22 − 16 = 6. The nth-term formula is aₙ = a₁ + (n − 1)d. Substituting a₁ = 4 and d = 6 gives aₙ = 4 + (n − 1)6 = 4 + 6n − 6 = 6n − 2. Therefore, option A is correct. Testing the rule, n = 1 gives 6 − 2 = 4, n = 2 gives 12 − 2 = 10, n = 3 gives 16, and n = 4 gives 22. Option B begins with 8, option C begins with 10, and option D has common difference 2, so they cannot describe the stated progression.

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.