nth term

This is a geometric sequence because each term is obtained by multiplying the previous term by 2. Here, the first term is \(a=7\) and the common ratio is \(r=2\). Therefore, \(a_n=ar^{n-1}\). So, \(a_{10}=7\times2^9=7\times512=3584\). The option 1792 uses \(2^8\), which gives the 9th term. Exam tip: In a geometric sequence, the exponent for the \(n\)th term is always \(n-1\).

Putting n=4, a_4=6(4)^2-6(4)+1=6×16-24+1=96-24+1=73. Therefore, the correct answer is 73. The value 72 may result from omitting the final +1, which is incorrect. Exam tip: Substitute the value of n in every term first, then simplify in order.

Observe that \(2=1^2+1\), \(5=2^2+1\), \(10=3^2+1\), \(17=4^2+1\), and \(26=5^2+1\). Therefore, the \(n\)th term is \(a_n=n^2+1\). The close distractor \(n^2+2\) gives 3 as its first term, so it does not fit. Exam tip: match the terms with squares of \(1,2,3,\ldots\) to identify a quadratic pattern quickly.

The governing concept is evaluating the same explicit sequence rule at two related indices and subtracting. First, aₙ₊₃ = 3(n + 3) + 2 = 3n + 9 + 2 = 3n + 11. Therefore aₙ₊₃ − aₙ = (3n + 11) − (3n + 2) = 9. Hence option A is correct. Another way to see this is that each increase of one in the index raises the term by 3; increasing the index by 3 raises it by 3 × 3 = 9. The distractors 6 and 3 reflect using the wrong index change or counting only one step, while 11 is the constant part of aₙ₊₃ rather than the required difference.

For an arithmetic sequence, the nth term is found from \\(a_n=a_1+(n-1)d\\), where \\(a_1\\) is the first term and \\(d\\) is the common difference. We are given \\(a_6=31\\) and \\(d=4\\). Substituting the sixth position gives \\(31=a_1+5(4)\\), so \\(a_1=31-20=11\\). Therefore, \\(a_n=11+(n-1)4=11+4n-4=4n+7\\). Hence option A is correct.

The check at the given position confirms the result: putting \\(n=6\\) into \\(4n+7\\) gives \\(4(6)+7=31\\), exactly as required. The expression \\(31+4n\\) incorrectly treats 31 as the first term and gives 55 at the sixth position. The other linear options also fail to preserve the stated sixth term or common difference. Thus the supplied answer A is mathematically consistent and the reasoning uses the correct relation for an arithmetic progression.

The term numbers differ by (2) and the common difference is (8), so the difference is (16). Use (d) in linear terms to save time.

To obtain the value 127, set \(5n^2+2=127\). This gives \(5n^2=125\), so \(n^2=25\) and \(n=5\). Since a term number is a positive integer, the correct answer is the 5th term. The 4th term is \(5\times4^2+2=82\), not 127. Exam tip: equate the given value to \(a_n\) first, then solve for \(n\).

Each term is multiplied by \(\frac{1}{2}\), so \(a_n=12\left(\frac{1}{2}\right)^{n-1}\). Keep the first term separate in a geometric sequence.

Given \(a_n=2n^2-3n+4\), replace \(n\) by \(n+1\): \(a_{n+1}=2(n+1)^2-3(n+1)+4=2n^2+n+3\). Therefore, \(a_{n+1}-a_n=(2n^2+n+3)-(2n^2-3n+4)=4n-1\). The option \(4n+1\) can result from an error while subtracting the constant terms. Exam tip: write \(a_{n+1}\) separately before subtracting \(a_n\).

Observe the terms in order: \(0=1^2-1\), \(3=2^2-1\), \(8=3^2-1\), \(15=4^2-1\), and \(24=5^2-1\). Therefore, the \(n\)th term is \(a_n=n^2-1\). For \(n^2+n\), the first term would be 2, so it does not match this sequence. Exam tip: substitute \(n=1\) and \(n=2\) to verify a proposed nth-term formula quickly.

To find the eighth term, substitute n=8: a_8=3(8)^2+8-2=3×64+8-2=192+8-2=198. Hence, 198 is correct. A value of 202 may result from incorrectly adding the final −2. Exam tip: evaluate the power first, then multiply, add, and subtract in order.

The terms match (1^2+4(1)+1,2^2+4(2)+1,\ldots). Testing options on the first three terms is a good method.

The first term is (9) and the difference is (7), so (a_n=9+(n-1)7=7n+2). In an arithmetic sequence, (n-1) differences are added.

To find the seventh term, substitute n=7: a_7=4(7)^2-3(7)+2=4×49-21+2=196-21+2=177. Therefore, 177 is the correct answer. A value such as 181 usually results from an error in subtraction or addition. Exam tip: evaluate the power first, then perform multiplication and addition/subtraction.

The governing concept is the nth-term formula for an arithmetic progression, aₙ = a₁ + (n − 1)d. The difference between the given terms is a₁₁ − a₅ = 64 − 28 = 36. Moving from the fifth term to the eleventh term takes six equal steps, so 6d = 36 and d = 6. Now use a₅ = a₁ + 4d: 28 = a₁ + 24, which gives a₁ = 4. Therefore aₙ = 4 + (n − 1)6 = 4 + 6n − 6 = 6n − 2. Thus option A is correct. Substitution confirms it: at n = 5 the value is 28 and at n = 11 it is 64.

Given \(a_n=7n-4\), substitute \(3m\) for \(n\): \(a_{3m}=7(3m)-4=21m-4\). Hence, the correct answer is \(21m-4\). The expression \(21m-12\) incorrectly multiplies the constant \(-4\) by 3. Exam tip: when the index changes, replace only \(n\) throughout the formula and do not alter a constant term unnecessarily.

The governing concept is the nth term of a geometric progression, aₙ = arⁿ⁻¹. Here the first term is a = 3 and the common ratio is r = 9/3 = 3. Therefore the eighth term is a₈ = 3 × 3⁷ = 3⁸. Evaluating the power gives 3⁸ = 6561. Hence option C, not option A, is correct. Option A, 2187, equals 3⁷ and results from using one fewer multiplication. Option B is 3⁶, while option D is 3⁵. The term number matters because the first term already contains one factor of 3, so the eighth term contains eight factors in total.

(a_3=4) and (a_8=34), so the correct sum is (38). Recheck calculation if options seem close.

The second differences are (4), and the terms match (2n^2+n+1). Equal second differences suggest a quadratic rule.

Given \(a_n=18-4n\), set the nth term equal to \(-30\): \(18-4n=-30\). Thus, \(-4n=-48\), so \(n=12\). Checking, \(a_{12}=18-4(12)=18-48=-30\). Hence, the 12th term is correct. The 13th term gives \(-34\), not \(-30\). Exam tip: handle signs carefully when solving equations involving negative numbers.