nth term

Given \(a_n=4n-3\), set the term equal to 45: \(4n-3=45\). Thus, \(4n=48\) and \(n=12\). Therefore, the 12th term is 45. The 11th term is \(4(11)-3=41\), so it is not correct. Exam tip: When a term value is given, equate \(a_n\) to that value and solve for \(n\).

This is an arithmetic progression with first term \(a=3\) and common difference \(d=4\). Its \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{15}=3+(15-1)\times4=3+56=59\). Hence, 59 is the correct option. Choosing 57 would add the common difference only 13 times, whereas the 15th term requires 14 differences. Exam tip: for the \(n\)th term, use \((n-1)\) common differences.

The first term is (5) and the ratio is (2), so (a_n=5\cdot2^{n-1}). In the geometric rule, the exponent is (n-1).

Given \(a_n=n^2+n\), substitute \(n=7\): \(a_7=7^2+7=49+7=56\). The value 54 would result from adding 5 to 49, so it is not correct. In exams, substitute the term number first and then evaluate the exponent.

The first term is (10) and the difference is (-3), so (a_n=10+(n-1)(-3)=13-3n). In a decreasing progression the difference is negative.

Given \(a_n=25-2n\), substitute \(n=10\): \(a_{10}=25-2(10)=25-20=5\). Hence, 5 is the correct option. The value 7 would be obtained for \(n=9\), so it is a close but incorrect distractor. Exam tip: after substituting the term number, multiply before subtracting.

This sequence is not arithmetic because its differences are 3, 5, and 7; instead, the terms follow a square-number pattern. Write each term using its position: 2 = 1² + 1, 5 = 2² + 1, 10 = 3² + 1, and 17 = 4² + 1. Therefore the general or nth-term rule is aₙ = n² + 1, so option A is correct. Substitution confirms it: for n = 1, a₁ = 2; for n = 2, a₂ = 5; for n = 3, a₃ = 10; and for n = 4, a₄ = 17. Option B is linear and gives 3 as the first term; option C gives 2, 6, 12, 20; option D gives 2 initially but then increases by 3 each time. Recognising the position-based square pattern is the key concept.

Given \(a_n=6n+1\), \(a_9=6(9)+1=55\) and \(a_4=6(4)+1=25\). Therefore, \(a_9-a_4=55-25=30\). Option 24 is not the correct difference when \(a_4=25\). Exam tip: You can also calculate directly: \(a_9-a_4=6(9-4)\).

The first term is \(a=3\) and the common ratio is \(r=\frac{12}{3}=4\). The nth term of a GP is \(a_n=ar^{n-1}\). Therefore, \(a_5=3\times4^{5-1}=3\times4^4=768\). Option 384 is a close distractor, but it equals \(3\times4^3\), which is the fourth term. Exam tip: in the nth-term formula, the exponent is always \(n-1\).

This is an arithmetic progression with first term \(a=8\) and common difference \(d=5\). Therefore, its \(n\)th term is \(a_n=a+(n-1)d=8+5(n-1)=5n+3\). Putting \(5n+3=58\) gives \(5n=55\), so \(n=11\). Hence, 58 is the 11th term. The 10th term is \(53\), so it is not correct. Exam tip: To find a term’s position, equate \(a_n\) to the given term.

Given \(a_n=2^n\), substitute \(n=6\): \(a_6=2^6=64\). The value 32 is \(2^5\), so it is a close but incorrect option. Exam tip: For an nth-term question, directly substitute the given term number for \(n\) in the formula.

The sequence follows a quadratic position pattern rather than a constant-difference pattern. Expressing the terms by their positions gives 5 = 1² + 4, 8 = 2² + 4, 13 = 3² + 4, and 20 = 4² + 4. Hence the nth term is aₙ = n² + 4, which is option A. Substitution verifies the rule for every displayed term: n = 1 gives 5, n = 2 gives 8, n = 3 gives 13, and n = 4 gives 20. The first differences are 3, 5, and 7, confirming that a simple arithmetic rule is not suitable. Option B is linear and gives 5, 7, 9, 11; option C gives 3, 8, 15, 24; option D gives 5 initially but then grows by 3. Only option A matches all terms.

For \(a_n=3n-2\), \(a_{n+1}-a_n=[3(n+1)-2]-(3n-2)=3\), which is constant for every \(n\). Hence it is an arithmetic progression. In \(n^2+1\), the differences change. Exam tip: compare consecutive terms’ difference.

This is an arithmetic progression with first term 12 and common difference 6. Therefore, \(a_{20}=a+(20-1)d=12+19\times6=126\). Getting 132 would mean adding 20 differences, but there are only 19 differences from the first term to the 20th term. Exam tip: in \(a_n=a+(n-1)d\), always use \(n-1\), not \(n\).

Given \(a_n=3^n\), substitute \(n=4\) to get \(a_4=3^4=81\). Although 27 is a power of 3, it equals \(3^3\), so it is the third term, not the fourth. Exam tip: To find a specified term from \(a_n\), substitute the subscript value for \(n\).

This is the sequence of consecutive odd numbers and forms an arithmetic progression with first term \(a=1\) and common difference \(d=2\). Therefore, \(a_n=a+(n-1)d=1+(n-1)\times2=2n-1\). For \(2n+1\), putting \(n=1\) gives 3, not the first term 1. Exam tip: always test an nth-term formula using \(n=1\).

To find the fourth term, substitute \(n=4\) in the rule: \(a_4=4(4)^2-1=4\times16-1=63\). Therefore, the correct answer is 63. A result of 62 would come from an error while subtracting the final 1. Exam tip: in expressions with powers, evaluate the square first, then multiply and subtract.

The common ratio is (\frac{1}{3}), and the terms are (81,27,9,3,1). In a decreasing progression, keep multiplying by the ratio.

This is an arithmetic progression with first term \(a=2\) and common difference \(d=2\). Therefore, \(a_n=a+(n-1)d=2+(n-1)\times2=2n\). Hence, the correct answer is \(2n\). The expression \(2n-1\) generates odd numbers, whereas every term in this sequence is even. Exam tip: always test a proposed nth-term formula at \(n=1\); \(2n\) gives the first term \(2\).

Given \(a_n=10n-4\). To find the eleventh term, substitute \(n=11\): \(a_{11}=10(11)-4=110-4=106\). Therefore, 106 is correct. The value 110 is obtained before subtracting 4, so it is a close but incorrect option. Exam tip: for the \(n\)th term, replace \(n\) with the required term number before simplifying.