nth term

This is an arithmetic progression because each term decreases by 4. Thus, the first term is \(a=30\) and the common difference is \(d=-4\). \(a_9=a+(9-1)d=30+8(-4)=30-32=-2\). Therefore, \(-2\) is the correct answer. The value \(0\) would result from subtracting the common difference only 7 times. Exam tip: for the \(n\)th term, use \(n-1\) common differences.

Given \(a_n=3n+14\) and \(a_n=56\), we get \(3n+14=56\). Subtracting 14 from both sides gives \(3n=42\), so \(n=14\). Hence, 56 is the 14th term of the sequence. If \(n=15\), the term would be \(3\times15+14=59\), not 56. Exam tip: To find the term number, substitute the given value of \(a_n\) into the formula and solve the resulting equation for \(n\).

The common ratio of this GP is \(r=\frac{48}{96}=\frac{1}{2}\). Therefore, the sixth term is \(a_6=96\left(\frac{1}{2}\right)^5=3\). The terms are \(96,48,24,12,6,3\), so 3 is correct. Option 6 is the fifth term, not the sixth. Exam tip: In \(a_n=a r^{n-1}\), use the exponent \(n-1\).

This is an arithmetic progression with first term \(a=9\) and common difference \(d=9\). Hence, \(a_n=a+(n-1)d=9+(n-1)\times9=9n\). The expression \(n+9\) gives 11 when \(n=2\), but the second term is 18. Exam tip: use \(a_n=a+(n-1)d\) for an arithmetic progression.

Given \(a_n=12-5n\), substitute \(n=4\) to find the fourth term: \(a_4=12-5(4)=12-20=-8\). Therefore, \(-8\) is correct. An option such as \(-6\) is incorrect because \(12-20=-8\), not \(-6\). Exam tip: after substituting the value of \(n\), do the multiplication before subtraction.

The consecutive differences are \(7,9,11\), increasing by \(2\) each time, so the sequence follows a quadratic pattern. Substituting \(n=1,2,3,4\) in \(a_n=n^2+4n\) gives \(5,12,21,32\), respectively. \(n^2+4\) matches only the first term and gives \(8\) for the second term. Exam tip: when first differences increase uniformly, test a quadratic nth-term rule.

Given \(a_n=3\cdot4^{n-1}\), substitute \(n=4\): \(a_4=3\cdot4^{4-1}=3\cdot4^3=3\cdot64=192\). Hence, 192 is correct. The value 96 may result from an incorrect multiplication after evaluating the power. Exam tip: first substitute the term number into \(n-1\), then simplify step by step.

This is an arithmetic progression with first term \(a=14\) and common difference \(d=4\). Therefore, \(a_n=a+(n-1)d=14+4(n-1)=4n+10\). Putting \(a_n=70\), we get \(4n+10=70\), so \(4n=60\) and \(n=15\). For \(n=14\), the term would be \(66\), so it is a close but incorrect option. Exam tip: to find the position of a term, first write \(a_n=a+(n-1)d\).

For the sixth term, substitute n=6: \(a_6=6^2+2(6)-1=36+12-1=47\). Therefore, the correct answer is 47. The value 49 would result from omitting the final \(-1\), so it is incorrect. Exam tip: Substitute the value of n in every occurrence of n and follow the order of operations carefully.

The first term of this GP is 2 and the common ratio is 5. Therefore, its nth term is \(a_n=2\times5^{n-1}\). From \(2\times5^{n-1}=1250\), we get \(5^{n-1}=625=5^4\), so \(n=5\). Hence, 1250 is the fifth term. The fourth term is \(250\), making it a close but incorrect option. Exam tip: To find a position in a GP, first write \(a_n=ar^{n-1}\).

Direct answer: aₙ = 10n − 4, so A is correct. The consecutive differences are 16 − 6 = 10, 26 − 16 = 10, and 36 − 26 = 10. This is an arithmetic sequence with first term a₁ = 6 and common difference d = 10. Apply aₙ = a₁ + (n − 1)d: aₙ = 6 + (n − 1)10 = 6 + 10n − 10 = 10n − 4. Check the first four positions: n = 1 gives 6, n = 2 gives 16, n = 3 gives 26, and n = 4 gives 36. B gives 14 at n = 1, so its first term is wrong. C gives 16 at n = 1 and increases by 6, not 10. D gives 11 at n = 1 and increases by only 1. A reliable method is to check both the starting value and the constant increase; either check alone may leave a misleading option.

Given \(a_n=\frac{5n-1}{2}\), substitute \(n=7\): \(a_7=\frac{5(7)-1}{2}=\frac{35-1}{2}=\frac{34}{2}=17\). Hence, 17 is correct. The value 16 could result from incorrectly calculating \(35-1\) as 32. Exam tip: In an nth-term formula, substitute the value of \(n\) carefully and simplify the operations in order.

The sequence follows a square-number pattern multiplied by 3. The first term is \(3\times1^2\), the second is \(3\times2^2\), the third is \(3\times3^2\), and the fourth is \(3\times4^2\). The differences are 9, 15, and 21, so this is not an arithmetic sequence with a constant difference; the square pattern is the useful observation.

Therefore, the nth term is \(a_n=3n^2\). Substitution verifies every displayed term: \(3(1)^2=3\), \(3(2)^2=12\), \(3(3)^2=27\), and \(3(4)^2=48\). Hence option A is correct. A formula such as \(3n+9\) is linear and cannot produce these values for all positions, even if it may match one term by chance.

The terms are (7,14,28,56,112,224,448), so (448) is the seventh term. In a geometric progression, multiply by the same ratio each time.

Substitute n=5: a_5=2(5)^2+3(5)=2×25+15=50+15=65. Therefore, 65 is the correct answer. An answer such as 60 usually results from an error in evaluating the square or multiplication. Exam tip: substitute the value first, then apply powers, multiplication, and addition in order.

The first term is (40) and the difference is (-5), so (a_n=40+(n-1)(-5)=45-5n). In a decreasing sequence, take the difference as negative.

Given \(a_n=\frac{n^2+n}{2}\). Substituting \(n=10\), \(a_{10}=\frac{10^2+10}{2}=\frac{100+10}{2}=\frac{110}{2}=55\). Hence, the correct answer is 55. Getting 50 is a common error caused by evaluating \(10^2\) incorrectly or not adding the numerator correctly. Exam tip: substitute the value of \(n\) first, then evaluate the power, addition, and division in order.

Given \(a_n=5\cdot2^{n-1}\), substitute \(n=6\): \(a_6=5\cdot2^{6-1}=5\cdot2^5=5\cdot32=160\). Hence, option C is correct. \(80\) would result from using exponent \(4\), but here \(n-1=5\). Exam tip: In an nth-term formula, substitute the value of \(n\) first and then evaluate the exponent carefully.

The direct answer is option B, \(6n+2\). This sequence is an arithmetic progression because the difference between neighbouring terms is constant: \(14-8=6\), \(20-14=6\), and \(26-20=6\). For an arithmetic progression, the nth term is \(a_n=a+(n-1)d\), where \(a\) is the first term and \(d\) is the common difference. Here, \(a=8\) and \(d=6\), so \(a_n=8+(n-1)6=8+6n-6=6n+2\). Checking: for \(n=1\), the expression gives 8; for \(n=2\), it gives 14; for \(n=3\), it gives 20. Therefore it matches every listed term. Option A, \(6n-2\), gives 4 when \(n=1\), so it starts incorrectly. Option B gives the correct first term and common increase. Option C, \(8n-6\), gives 2, 10, 18, so its difference is 8, not 6. Option D, \(2n+6\), gives 8 initially but then increases by only 2, giving 10 next instead of 14. Memory cue: first term plus \((n-1)\) times the common difference.

To find the 10th term, substitute 10 for n in the given rule: \(a_{10}=5\times10+4=50+4=54\). Therefore, 54 is the correct option. The value 52 would result if the constant term were 2, which it is not. Exam tip: While finding an nth term, substitute the term number carefully for n.

For an arithmetic progression, the nth term is a_n = a + (n − 1)d. Here a = 11, d = 3, and n = 16, so a_16 = 11 + (16 − 1)×3 = 11 + 45 = 56. Thus option B is correct. The common mistake is multiplying d by 16 instead of by n − 1, which would produce an incorrect distractor.

In this geometric progression, the first term is \(a=6\) and the common ratio is \(r=\frac{12}{6}=2\). The \(n\)th term is \(a_n=ar^{n-1}\). Hence, \(a_7=6\times2^{7-1}=6\times64=384\). Option 192 equals \(6\times2^5\), so it is the sixth term. Exam tip: in a GP, the exponent in the nth-term formula is \(n-1\), not \(n\).

The consecutive differences are \(3,5,7,\ldots\), which are odd numbers. This indicates a pattern involving \(n^2\). Substituting \(n=1,2,3,4\) in \(n^2+5\) gives \(6,9,14,21\), respectively. Although \(3n+3\) gives the first two terms, it gives \(12\) as the third term, so it is incorrect. Exam tip: when the differences are consecutive odd numbers, test an nth-term rule involving \(n^2\).

Substitute n=5: a_5=4(5)^2+3=4×25+3=103. Therefore, 103 is correct. A value such as 99 may result from evaluating the square incorrectly or making an arithmetic error. In exams, substitute the value of n first, then evaluate the power, multiplication, and addition in order.

The governing concept is using the general term of an arithmetic progression to find the position of a given value. The first term is a₁ = 7 and the common difference is d = 13 − 7 = 6. Therefore, aₙ = a₁ + (n − 1)d = 7 + 6(n − 1) = 6n + 1. To locate 61, set 6n + 1 = 61. Subtracting 1 from both sides gives 6n = 60, and dividing by 6 gives n = 10. Hence option C is correct: 61 is the tenth term. Checking directly, the tenth term is 7 + 9×6 = 61. The other choices arise from errors in counting terms or solving the linear equation.