nth term

For the second term, substitute n=2. Thus, a_2=5^2=25, so 25 is correct. The value 125 equals 5^3, so it is the third term. Exam tip: carefully substitute the required term number for n in the nth-term formula.

The formula gives the value of a term when its position number is known. For the fourth term, the position is represented by putting \\(n=4\\) into the given rule. The expression has a product in the numerator and division by 2, so both the multiplication and the division must be completed carefully. This type of formula is useful because it finds a term directly without listing all earlier terms.

Substitute 4: \\(a_4=\\frac{4(4+1)}{2}=\\frac{4\\times5}{2}=\\frac{20}{2}=10\\). Therefore, the fourth term is 10, so option D is correct. The value 6 would result from using an unsuitable expression, while 8 or 9 do not follow from the stated formula. The key step is to use the position number 4, not the term value from some other pattern.

Its rule is (a_n=n(n+1)), and (6\times7=42). In exams, match the given term with the rule to find the term number.

The first term is (8) and the common difference is (6), so (a_n=8+(n-1)6=6n+2). Check the rule by putting (n=1).

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 the general term of an arithmetic progression. The first term is a = 7 and the common difference is d = 13 − 7 = 6. Hence the nth term is a_n = a + (n − 1)d = 7 + 6(n − 1) = 6n + 1. To find the position of 61, set 6n + 1 = 61. Subtracting 1 gives 6n = 60, and dividing by 6 gives n = 10. Therefore, option C is correct: 61 is the tenth term. The other choices result from an incorrect difference or an arithmetic error in solving the equation.

Given \(a_n=7n-1\), set the term equal to 69: \(7n-1=69\). Thus, \(7n=70\), so \(n=10\). Hence, 69 is the 10th term of the sequence. The 9th term is \(7\times9-1=62\), so it is not correct. Exam tip: To find the position of a given term, equate \(a_n\) to that value and solve for \(n\).

This is an arithmetic progression with first term 5 and common difference 7. Therefore, \(a_{13}=a+(13-1)d=5+12\times7=89\). Hence, the correct answer is 89. A value such as 87 can result from incorrectly counting the number of common differences. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

The first term is (7) and the ratio is (3), so (a_n=7\cdot3^{n-1}). In the geometric rule, the exponent is (n-1).

To find the seventh term, substitute \(n=7\) in the formula: \(a_7=7^2+3\times7=49+21=70\). Therefore, the correct answer is 70. Getting 77 may result from an error in multiplication or addition. Exam tip: after substituting the term number, evaluate powers and multiplication before addition.

The first term is (21) and the difference is (-5), so (a_n=21+(n-1)(-5)=26-5n). In a decreasing sequence, the difference is negative.

Given \(a_n=55-4n\), substitute \(n=11\): \(a_{11}=55-4(11)=55-44=11\). Hence, 11 is the correct answer. Getting 9 would require an incorrect multiplication or subtraction. Exam tip: after substituting the value of \(n\), multiply first and then subtract.

Given \(a_n=10n+1\), \(a_8=10\times8+1=81\) and \(a_3=10\times3+1=31\). Therefore, \(a_8-a_3=81-31=50\). Option 45 may result from using only the index difference \(8-3\), but each increase of 1 in the index increases the term by 10. Exam tip: find both terms, or directly use \(a_8-a_3=10(8-3)\).

This is an arithmetic progression with first term 10 and common difference 7. Its nth term is \(a_n=10+(n-1)\times 7=7n+3\). Setting \(7n+3=73\) gives \(7n=70\), so \(n=10\). Therefore, 73 is the 10th term. The 9th term is 66, so it is a close but incorrect option. Exam tip: To find the position of a given term, form \(a_n\) first and equate it to that term.

Given \(a_n=6^n\), substitute \(n=3\) to find the third term: \(a_3=6^3=6\times6\times6=216\). Hence, 216 is correct. The value 36 is \(6^2\), so it is not the third term. Exam tip: To find an \(n\)th term, substitute the required value of \(n\) carefully into the formula.

The correct term is \(a_n=n^2+2n+5\). Substituting \(n=1,2,3,4\) gives \(8,13,20,29\), respectively. In option B, the second term is \(11\), while in option D it is \(14\), so neither matches the sequence. Exam tip: For sequences with changing first differences, verify a possible quadratic expression using the first few values of \(n\).