nth term

The successive differences are \(17,25,33,41\), and their second differences are all \(8\). Hence the nth term is quadratic: \(a_n=4n^2+bn+c\). Substituting \(n=1\) and \(n=2\) gives \(b=5\) and \(c=0\). Therefore, \(a_n=4n^2+5n\). Option C gives 9 for \(n=1\), but it gives 25 rather than 26 for \(n=2\). Exam tip: when second differences are constant, test a form \(an^2+bn+c\).

The governing idea is substitution into an explicit general term. The formula aₙ = 11n − 15 is valid for any permitted index, so replace the index n by the entire expression 2r − 1. Thus a₂ᵣ₋₁ = 11(2r − 1) − 15. Distributing 11 gives 22r − 11 − 15, and combining constants gives 22r − 26. Therefore option A is correct. The complete index must be substituted; replacing it only by r would produce the wrong expression. Option B forgets to subtract the contribution from −1, option C uses 11r instead of 22r, and option D adds rather than correctly combines the constant terms.

This is a geometric progression because each term is obtained by multiplying the preceding term by 5: 30/6 = 5, 150/30 = 5, and 750/150 = 5. For a geometric sequence, aₙ = arⁿ⁻¹, where a is the first term and r is the common ratio. Here a = 6 and r = 5, so a₆ = 6 × 5⁵. Since 5⁵ = 3125, a₆ = 6 × 3125 = 18750. Thus option A is correct. The exponent is 5 rather than 6 because the first term already contains the initial factor. Options C and D use an insufficient or incorrect multiplication, while option B does not follow the progression.

Given \(a_n=2n^2+9n-11\), \(a_{10}=2(10)^2+9(10)-11=279\) and \(a_4=2(4)^2+9(4)-11=57\). Therefore, \(a_{10}-a_4=279-57=222\). The value 204 results from incorrectly calculating \(a_4\) as 75. Exam tip: substitute each value of \(n\) separately before finding the difference.

The second differences are (8), and (4n^2+1) gives all starting terms. Half of the second difference is the coefficient of (n^2).

Given \(a_n=35-6n\), put \(a_n=-43\): \(35-6n=-43\). Thus, \(-6n=-78\), so \(n=13\). Therefore, \(-43\) is the 13th term. The nearby distractor, the 14th term, is incorrect because \(a_{14}=35-84=-49\). Exam tip: take special care with signs when transposing negative terms in an equation.

The sequence has first differences 13, 19, 25, and 31; these increase by 6, so a quadratic formula is appropriate. Test option C at the first few indices: for n = 1, 3(1)² + 4(1) + 1 = 8; for n = 2, 3(4) + 8 + 1 = 21; for n = 3, 3(9) + 12 + 1 = 40. It also gives n = 4: 48 + 16 + 1 = 65, and n = 5: 75 + 20 + 1 = 96. Hence aₙ = 3n² + 4n + 1 and option C is correct. The other formulas fail at least one early term, so checking several indices removes accidental matches.

For (n=9), ((-1)^9=-1) and (5(9)-3=42), so (a_9=-42). In alternating signs, check the even-odd position.

The governing concept is the nth-term formula for an arithmetic progression: aₙ = a₁ + (n − 1)d. Since the ninth term is 61 and the common difference is 7, move backward eight common differences to obtain the first term: a₁ = 61 − 8(7) = 61 − 56 = 5. Therefore, aₙ = 5 + (n − 1)7 = 5 + 7n − 7 = 7n − 2. Substitution of n = 9 gives 63 − 2 = 61, confirming the result. Option B incorrectly uses a different constant, option C shifts it still further, and option D has the wrong coefficient of n.

The governing idea is to express each term using the same power pattern. The sequence can be rewritten as 13 = 16 − 3, 29 = 32 − 3, 61 = 64 − 3, 125 = 128 − 3, and 253 = 256 − 3. These powers are 2⁴, 2⁵, 2⁶, 2⁷, and 2⁸, so the power in the nth term is 2ⁿ⁺³. Hence aₙ = 2ⁿ⁺³ − 3. For n = 1, this gives 16 − 3 = 13, and for n = 5 it gives 256 − 3 = 253. Option A gives the wrong exponent and constant, option C does not fit the first term, and option D is linear rather than exponential.

Each term increases by (\frac{11}{6}), so (a_n=\frac{11n}{6}). Use a common denominator to see the pattern in fractions.

The terms match (5n^2+1), so none of the listed options is correct. Always test options on the first term.

Given \(a_n=8n^2-7n+4\), substitute \(n+1\) for \(n\): \(a_{n+1}=8(n+1)^2-7(n+1)+4=8n^2+9n+5\). Therefore, \(a_{n+1}-a_n=(8n^2+9n+5)-(8n^2-7n+4)=16n+1\). The option \(16n-7\) results from incorrectly handling the linear term produced when expanding \((n+1)^2\). Exam tip: find \(a_{n+1}\) completely before subtracting \(a_n\).

Given \(a_n=2n^3-3n\), substitute \(n=5\): \(a_5=2(5)^3-3(5)=2\times125-15=250-15=235\). Hence, 235 is correct. 250 is only the value of \(2(5)^3\); the term \(3\times5\) must still be subtracted. Exam tip: evaluate the power first, then multiply and subtract.

The first term is (44) and the difference is (-7), so (a_n=44+(n-1)(-7)=51-7n). Use a negative difference in a decreasing sequence.

For an AP, \(a_n=a_1+(n-1)d\). Substituting \(a_1=11\) and \(d=-4\) gives \(a_n=11-4(n-1)\). In option D, the first term is 7, not 11. Exam tip: always verify the first term first.

The second differences are (4), and (2n^2+10n+6) gives the starting terms. Confirm a quadratic rule using the first three terms.

Given \(a_n=13n-19\), set the nth term equal to 176: \(13n-19=176\). Thus, \(13n=195\), so \(n=15\). Therefore, 176 is the 15th term of the sequence. The 14th term is \(13(14)-19=163\), so it is not correct. Exam tip: To find a term number, equate \(a_n\) to the given term and solve for n.

Each term increases by (\frac{13}{7}), so (a_n=\frac{13n}{7}). In fractions with a common denominator, observe the numerator pattern.

Substituting the values in the given formula gives \(a_9=3(9)^2+5(9)-8=280\) and \(a_6=3(6)^2+5(6)-8=130\). Therefore, \(a_9-a_6=280-130=150\). Option 120 is incorrect because it uses 160 as the value of \(a_6\). Exam tip: Evaluate both terms separately before finding their difference.