nth term

The direct answer is A: \(4\cdot3^{n-1}\). A geometric progression is a sequence in which each term is obtained by multiplying the previous term by the same number. Here the first term is \(a=4\), and the common ratio is \(r=12/4=3\); also \(36/12=3\). The nth-term rule is \(a_n=ar^{n-1}\), because the first term has zero multiplications by the ratio, the second has one, and the nth term has \(n-1\) multiplications. Thus \(a_n=4\cdot3^{n-1}\). Check: for \(n=1\), it gives 4; for \(n=2\), it gives 12; for \(n=3\), it gives 36. Option A has the correct first term, ratio, and exponent. Option B uses 3 as the first term and 4 as the ratio, reversing the roles. Option C, \(4n+3\), is a linear expression and does not produce 4, 12, 36. Option D starts with 12, so at \(n=1\) it gives 12 instead of 4. Exam cue: in a GP, identify the first term and common ratio before using \(ar^{n-1}\).

Substitute 8 for n: a_8=8^2+2(8)=64+16=80. Therefore, 80 is the correct answer. The value 84 may result from an addition error. Exam tip: When finding a term, substitute the value of n everywhere in the formula before evaluating powers and multiplication.

Given (a_n=40-3n), substitute n=12 to find the 12th term: (a_{12}=40-3(12)=40-36=4). Hence, the correct answer is 4. The value 6 can result from an incorrect multiplication or subtraction. In exams, multiply first and then subtract.

The terms are \(2^2,3^2,4^2,5^2\), respectively. For \(n=1\), the first term is \((1+1)^2=4\); hence the \(n\)th term is \((n+1)^2\). Although \(n^2+3\) gives 4 for the first term, it gives 7 when \(n=2\), not the required second term 9. Exam tip: verify a proposed formula using at least the first two terms.

Given \(a_n=8n+3\), \(a_7=8(7)+3=59\) and \(a_2=8(2)+3=19\). Therefore, \(a_7-a_2=59-19=40\). Option 36 may result from an error in multiplication or subtraction. Exam tip: for questions involving two terms, substitute each value of \(n\) separately before finding their difference.

In this geometric progression, the first term is \(a=2\) and the common ratio is \(r=8/2=4\). The \(n\)th term is \(a_n=a r^{n-1}\). Therefore, \(a_5=2\times4^{5-1}=2\times4^4=512\). Choosing 256 results from using one power too few. Exam tip: in the \(n\)th-term formula, the exponent of the common ratio is always \(n-1\).

This is an arithmetic progression with first term \(a=3\) and common difference \(d=7\). Its \(n\)th term is \(a_n=a+(n-1)d=3+7(n-1)=7n-4\). Setting \(7n-4=80\) gives \(7n=84\), so \(n=12\). Hence, 80 is the 12th term. The 11th term is \(73\), so it is not correct despite being a close option. Exam tip: When the position of a given term is asked, equate \(a_n\) to that term and solve for \(n\).

Given \(a_n=4^n\), substitute \(n=3\): \(a_3=4^3=4\times4\times4=64\). Option 16 is the value of \(4^2\), not \(4^3\). Exam tip: For an nth-term question, first substitute the required term number for \(n\), then evaluate the power.

The successive differences are \(4,6,8\), increasing by \(2\) each time, so the rule should be quadratic. Substituting \(n=1,2,3,4\) in \(n^2+n+4\) gives \(6,10,16,24\), respectively. In contrast, \(n^2+5\) gives \(9\) as the second term, so it does not fit. Exam tip: when second differences are constant, test a rule containing \(n^2\).

Given \(a_n=9n-2\), \(a_5=9\times5-2=43\) and \(a_3=9\times3-2=25\). Therefore, \(a_5+a_3=43+25=68\). Option 66 can result from an error while subtracting in one of the terms. Exam tip: Substitute the value of \(n\) separately for each required term before adding.

This is an arithmetic progression with first term \(a=13\) and common difference \(d=19-13=6\). Its \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{18}=13+(18-1)\times6=13+102=115\). Option 109 uses 16 common differences, so it is the 17th term. Exam tip: Remember to use \(n-1\), not \(n\), in the nth-term formula.

Given \(a_n=5^n\). To find the third term, substitute \(n=3\): \(a_3=5^3=125\). The value 25 is \(5^2\), so it represents the second term, not the third. Exam tip: the subscript \(n\) indicates the term number; substitute it into the formula.

These are odd numbers starting from (3), so (a_n=2n+1). For (n=1), the first term should be (3).

Given \(a_n=5n^2-2\), substitute \(n=3\) for the third term: \(a_3=5(3)^2-2=5\times9-2=45-2=43\). Therefore, 43 is correct. The close distractor 45 results from forgetting to subtract 2 at the end. Exam tip: after substituting the value of \(n\), evaluate the exponent first, then multiply and subtract.

The common ratio is \(\frac{1}{2}\), and \(a_5=100\left(\frac{1}{2}\right)^4=\frac{25}{4}\). Use powers carefully with a fractional ratio.

This is an arithmetic progression with first term \(a=4\) and common difference \(d=4\). Hence, \(a_n=a+(n-1)d=4+(n-1)\times4=4n\). The expression \(4n-4\) gives 0 when \(n=1\), but the first term is 4. Exam tip: For such sequences, use \(a_n=a+(n-1)d\) and verify it with the first term.

Given \(a_n=12n-7\), substitute \(n=10\) to find the tenth term: \(a_{10}=12\times10-7=120-7=113\). Hence, 113 is correct. The value 117 would result from incorrectly adding 7 instead of subtracting it. Exam tip: in \(a_n\), replace \(n\) with the number of the required term.

The first term is (25) and the difference is (-5), so (a_n=25+(n-1)(-5)=30-5n). Pay attention to the sign in a decreasing sequence.

Given \(a_n=n(n+2)\), substitute \(n=5\) to find the fifth term: \(a_5=5(5+2)=5\times7=35\). Hence, 35 is correct. The value 30 would result from not multiplying 5 by \(5+2\) correctly. Exam tip: For an nth-term formula, substitute the given value of \(n\) first and then simplify the brackets.

The direct answer is option C, 162. This is a geometric progression because the same multiplier is used each time. The common ratio is \(r=18\div6=3\), and \(54\div18=3\). The formula is \(a_n=ar^{n-1}\). For the fourth term, use \(a=6\), \(r=3\), and \(n=4\): \(a_4=6\times3^{3}=6\times27=162\). We can also continue the pattern directly: 6, 18, 54, then \(54\times3=162\). Option A, 108, is not the result of multiplying 54 by 3 and does not fit the progression. Option B, 144, also does not follow the common ratio. Option C, 162, is correct because it is the next term after 54. Option D, 216, uses an incorrect multiplier or counting step. The key memory rule is that the exponent in the nth-term formula is one less than the term number.

Given \(a_n=7n+4\), we get \(a_8=7\times8+4=60\) and \(a_5=7\times5+4=39\). Hence, \(a_8-a_5=60-39=21\). Choosing 28 would result from using the difference in indices incorrectly. Exam tip: for \(a_n=dn+c\), \(a_p-a_q=d(p-q)\), since the constant term \(c\) cancels out.

The successive differences are \(4,6,8,\ldots\), increasing by 2 each time. Hence the sequence follows a quadratic rule. Substituting \(n=1,2,3,4\) in \(a_n=n^2+n+5\) gives \(7,11,17,25\), respectively. The option \(4n+3\) has a constant difference of 4, so it cannot represent this sequence. Exam tip: increasing first differences usually indicate a quadratic sequence.

To find the zero term, set \(a_n=0\) in the given expression: \(60-6n=0\). Thus, \(6n=60\), so \(n=10\). Therefore, the 10th term is zero. The 9th term is not zero because \(a_9=60-6(9)=6\). Exam tip: When asked for a zero term, first put \(a_n=0\) and solve for \(n\).

The terms are (5,15,45,135,405,1215), so (1215) is the sixth term. For small questions, listing terms is quick.

To find \(a_6\), substitute \(n=6\) in the formula: \(a_6=\frac{6(6+3)}{2}=\frac{6\times9}{2}=27\). Therefore, the correct answer is 27. The value 30 can result from incorrectly taking \(6+3\) as 10. Exam tip: In nth-term questions, substitute the given value of \(n\) carefully before simplifying.