nth term

The direct answer is option B, 500. This is a geometric progression because each term is obtained by multiplying the previous term by the same number. The common ratio is \(r=20\div4=5\), and \(100\div20=5\), so the pattern is reliable. The nth-term rule is \(a_n=ar^{n-1}\). Here \(a=4\), \(r=5\), and \(n=4\). Therefore \(a_4=4\times5^{4-1}=4\times5^3=4\times125=500\). Option A, 400, is wrong because it does not follow the required multiplication pattern. Option B, 500, is correct because it is the fourth term. Option C, 600, is not obtained by multiplying the first term by the correct power of 5. Option D, 800, is also too large and results from an incorrect calculation. Remember: in a geometric progression, the first term is multiplied by the ratio \(n-1\) times, not \(n\) times.

Given \(a_n=5n+6\), \(a_7=5\times7+6=41\) and \(a_3=5\times3+6=21\). Therefore, \(a_7-a_3=41-21=20\). The value \(25\) would result if the difference between the term numbers were 5, but here \(7-3=4\). Exam tip: for a linear sequence \(a_n=dn+c\), use \(a_p-a_q=d(p-q)\) to find the difference quickly.

Compare the terms in order: \(4=1^2+3\), \(7=2^2+3\), \(12=3^2+3\), and \(19=4^2+3\). Therefore, the \(n\)th term is \(a_n=n^2+3\). The option \(n^2+2\) is close, but it gives \(3\) as the first term instead of \(4\). Exam tip: for such sequences, compare the first few terms with \(1^2,2^2,3^2,\ldots\) and check for a constant difference.

For an AP, \(a_n=a+(n-1)d\). Here \(a=20\) and \(d=-3\), so \(a_n=20-3(n-1)\). Formula \(20-3n\) gives 17 as the first term. Exam tip: substitute \(n=1\) to check a proposed nth-term formula.

Direct answer: Option C, the sixth term, is correct. The sequence starts at 9 and each term is multiplied by 2: 9, 18, 36, 72, 144, 288. Count positions rather than only counting multiplications: 9 is first, 18 second, 36 third, 72 fourth, 144 fifth, and 288 sixth. Algebraically, the nth term is \\(a_n=9\cdot2^{n-1}\\). Setting it equal to 288 gives \\(9\cdot2^{n-1}=288\\), so \\(2^{n-1}=32=2^5\\), hence \\(n-1=5\\) and \\(n=6\\). Option A is wrong because the fourth term is 72. B is wrong because the fifth is 144. C matches the calculation. D would be the next term, 576. Memory cue: write a few terms and number them before selecting.

Given \(a_n=\frac{n(n+1)}{2}\). Substituting \(n=8\), we get \(a_8=\frac{8(8+1)}{2}=\frac{8\times9}{2}=36\). Therefore, option C is correct. A value such as \(40\) can result from substituting incorrectly in the formula. Exam tip: replace \(n\) with the given term number before calculating.

This is an arithmetic progression with first term \(a=20\) and common difference \(d=17-20=-3\). Its \(n\)th term is \(a_n=a+(n-1)d\). Hence, \(a_8=20+(8-1)(-3)=20-21=-1\). Therefore, -1 is correct. The value 0 can result from incorrectly using the term number or forgetting the \((n-1)\) factor. Exam tip: In \(a_n\), add the common difference \((n-1)\) times, not \(n\) times.

Given \(a_n=3n^2-2n\), substitute \(n=4\): \(a_4=3(4)^2-2(4)=3\times16-8=48-8=40\). Therefore, 40 is correct. The value 36 can result from an error while evaluating \(2\times4\) or during subtraction. Exam tip: substitute the term number first, then follow the order of powers, multiplication, and subtraction.

Given \(a_n=2n+9\) and \(a_n=35\), we get \(2n+9=35\). Subtracting 9 gives \(2n=26\), so \(n=13\). Thus, 35 is the 13th term of the sequence. For the close distractor 12, \(a_{12}=2(12)+9=33\), not 35. Exam tip: To find a term’s position, substitute its value for \(a_n\) and solve the resulting equation.

This is an arithmetic progression with first term \(a=7\) and common difference \(d=7\). Therefore, \(a_n=a+(n-1)d=7+(n-1)\times7=7n\). Substituting \(n=1\) gives \(7\), and substituting \(n=4\) gives \(28\), so \(7n\) is correct. The expression \(n+7\) increases by only 1 each time, so it does not represent this sequence. Exam tip: use \(a_n=a+(n-1)d\) for an arithmetic progression.

To find the fifth term, substitute n=5 in the formula: \(a_5=9-4(5)=9-20=-11\). Therefore, the correct answer is -11. The value -9 would result from incorrectly subtracting 18 from 9 instead of calculating 4×5 correctly. Exam tip: For an nth-term formula, substitute the value of n first, multiply, and then perform addition or subtraction.

The successive differences are 5, 7, and 9, which are consecutive odd numbers, so the pattern is quadratic. Substituting \(n=1,2,3,4\) into \(n^2+2n\) gives 3, 8, 15, and 24 respectively. Therefore, option A is correct. Option B gives 6 when \(n=2\), not 8. Exam tip: verify an nth-term formula by checking it against the first few terms.

Substituting \(n=4\), \(a_4=2\cdot5^{4-1}=2\cdot5^3=2\cdot125=250\). Therefore, the correct answer is 250. The exponent is \(n-1\), so it must be evaluated as \(4-1=3\); using \(5^4\) would give an incorrect value. Exam tip: substitute the term number into the exponent carefully before calculating.

This is an arithmetic sequence with first term 11 and common difference 4. Therefore, \(a_n=11+(n-1)\times4=4n+7\). Substituting \(a_n=63\) gives \(4n+7=63\), so \(4n=56\) and \(n=14\). Hence, 14 is the correct option. The 15th term is 67, so 15 is a close but incorrect distractor. Exam tip: To find a term position, substitute the given term in the formula for \(a_n\) and solve for \(n\).

Given (a_n=n^2-1), substitute n=9 to find the ninth term: (a_9=9^2-1=81-1=80). Therefore, 80 is correct. The value 81 is only the square of 9; subtracting 1 is still required. In exams, substitute the term number first and then follow the order of operations carefully.

In this GP, the first term is \(a=1\) and the common ratio is \(r=4\). Therefore, the \(n\)th term is \(T_n=ar^{n-1}=4^{n-1}\). Since \(1024=4^5\), we get \(n-1=5\), so \(n=6\). Hence, 1024 is the sixth term. The fifth term is \(4^4=256\), not 1024. Exam tip: In a GP, compare the exponent with \(n-1\) to find the term position.

The governing concept is the nth term of an arithmetic progression. The first term is a₁ = 5, and the common difference is d = 15 − 5 = 10. For an arithmetic progression, aₙ = a₁ + (n − 1)d. Substituting the known values gives aₙ = 5 + (n − 1)10 = 5 + 10n − 10 = 10n − 5. Therefore option A is correct. Checking the rule gives 5, 15, 25, and 35 for n = 1, 2, 3, and 4. Option B gives 15 for the first term, option C also gives 15, and option D gives −5, so these alternatives fail the first-term test.

To find the seventh term, substitute \(n=7\): \(a_7=\frac{3(7)+1}{2}=\frac{22}{2}=11\). Therefore, the correct answer is 11. Option 10 could result from omitting the addition of 1 in the numerator. Exam tip: after substituting the value of \(n\), use brackets and follow the order of operations carefully.

The sequence is not arithmetic because its differences are 6, 10, and 14, which are not constant. Instead, compare each term with the square of its position: the first term is 2 times \(1^2\), the second is 2 times \(2^2\), the third is 2 times \(3^2\), and the fourth is 2 times \(4^2\). This reveals the general pattern.

Thus the nth term is \(a_n=2n^2\). Checking the formula gives \(2(1)^2=2\), \(2(2)^2=8\), \(2(3)^2=18\), and \(2(4)^2=32\), exactly matching the given terms. Therefore, option A is correct. The linear choices cannot produce the increasing changes seen in this sequence.

For the ninth term, substitute 9 for n in the given rule: \(a_9=4\times 9+7=36+7=43\). Therefore, 43 is the correct answer. The value 41 would result from an error while adding 7. Exam tip: after substituting the term number in an nth-term formula, perform multiplication before addition.

For \(a_n=3n+1\), \(a_{n+1}-a_n=[3(n+1)+1]-(3n+1)=3\), a constant. Hence it is an AP. For \(n^2+1\), the differences change. Exam tip: a linear form \(pn+q\) represents an AP.

In this geometric progression, the first term is \(a=3\) and the common ratio is \(r=9/3=3\). Using \(a_n=ar^{n-1}\), \(a_5=3\times 3^{4}=243\). Here, 81 is the fourth term, while 729 is the sixth term. Exam tip: for the \(n\)th term, the exponent of the common ratio is always \(n-1\).

Observe the terms: \(3=1^2+2\), \(6=2^2+2\), \(11=3^2+2\), and \(18=4^2+2\). Hence, the \(n\)th term is \(a_n=n^2+2\). The expression \(2n+1\) would give a constant difference of \(2\), but this sequence has differences \(3,5,7,\ldots\). Exam tip: sequences based on squares have consecutive differences that are odd numbers.

The direct answer is option C, the 10th term. The sequence increases by 5 each time: \(17-12=5\), \(22-17=5\), and \(27-22=5\). Thus it is an arithmetic progression with first term \(a=12\) and common difference \(d=5\). Its nth term is \(a_n=a+(n-1)d=12+5(n-1)=5n+7\). To find the position of 57, set the term formula equal to 57: \(5n+7=57\). Subtract 7: \(5n=50\); divide by 5: \(n=10\). Option A, 8th, would give \(12+7\times5=47\), not 57. Option B, 9th, gives 52. Option C, 10th, gives 57 and is correct. Option D, 11th, gives 62. The exam cue is simple: for an arithmetic sequence, write the first term plus \((n-1)\) times the common difference.

This is an arithmetic progression because the difference between consecutive terms is \(7\). Here, \(a=2\), \(d=7\), and \(n=14\). Therefore, \(a_{14}=a+(n-1)d=2+(14-1)\times7=2+91=93\). Option 91 uses only \(13\times7\) and does not add the first term, 2. Exam tip: For the \(n\)th term, use \(n-1\) common differences, not \(n\).