nth term

Find the difference between consecutive terms: \(a_{n+1}-a_n=[7(n+1)+4]-(7n+4)=7\). Since this difference is constant, the sequence is an arithmetic progression with common difference 7. It is not a geometric progression because the ratio of consecutive terms is not constant. Exam tip: for a sequence of the form \(a_n=pn+q\), the common difference is \(p\).

Given \(a_n=n^2\), the previous term is \(a_{n-1}=(n-1)^2\). Therefore, \(a_n-a_{n-1}=n^2-(n-1)^2=n^2-(n^2-2n+1)=2n-1\). The expression \(2n+1\) is obtained from \((n+1)^2-n^2\), so it is not correct here. Exam tip: To find \(a_{n-1}\), replace every \(n\) by \(n-1\).

For consecutive terms, \(a_{n+1}-a_n=[7(n+1)+1]-(7n+1)=7\). Since this difference is constant, the sequence is an arithmetic progression with common difference 7. Option B is incorrect because a geometric progression must have a constant ratio between consecutive terms. Exam tip: for a sequence of the form \(a_n=dn+c\), \(d\) is the common difference.

The consecutive differences are 3, 5, 7, and 9, which are successive odd numbers. This indicates a rule involving \(n^2\). To obtain the first term 3 when \(n=1\), add 2 to \(1^2\); hence \(a_n=n^2+2\). Checking: for \(n=2\), it gives 6, and for \(n=3\), it gives 11. Option A gives 2 as the first term, so it is incorrect. Exam tip: For a quadratic sequence, inspect the first and second differences to identify the rule.

For 124, set \(5n^2-1=124\). Then \(5n^2=125\), so \(n^2=25\) and \(n=5\). Since a term number must be a positive integer, 124 is the fifth term. Although \(n=-5\) can arise algebraically from squaring, it cannot be a term number. Exam tip: To find the position of a given term, equate \(a_n\) to that value and solve for \(n\).

Substitute \(n=8\) in \(a_n=\frac{2n+1}{3}\): \(a_8=\frac{2(8)+1}{3}=\frac{16+1}{3}=\frac{17}{3}\). Hence, \(\frac{17}{3}\) is correct. The value \(\frac{16}{3}\) would result from incorrectly omitting the \(+1\) in the numerator. Exam tip: substitute the given value of \(n\) into every part of the formula before simplifying.

Given \(a_n=3n-8\), \(a_1=3(1)-8=-5\) and \(a_{15}=3(15)-8=37\). Therefore, \(a_1+a_{15}=-5+37=32\). Option \(37\) is only the 15th term, not the required sum. Exam tip: substitute each value of \(n\) separately before adding the terms.

The consecutive terms satisfy \(a_{n+1}=2a_n+4\): \(4\to12\to28\to60\). The corresponding nth term is \(a_n=2^{n+2}-4\). Checking, for \(n=1\), it gives \(8-4=4\), and for \(n=2\), it gives \(16-4=12\). The option \(2^{n+2}+4\) gives 12 as the first term, so it is incorrect. Exam tip: substitute the first two values of \(n\) to check a proposed sequence formula quickly.

For positive integer values of n, \(a_n=n^2-3n+5=(n-1)(n-2)+3\). The product \((n-1)(n-2)\) is 0 at n = 1 and n = 2, and it is positive for other positive integers. Hence, the smallest term value is \(3\). Option \(2\) may seem tempting because the quadratic has its real minimum at \(n=\tfrac32\), but a sequence index must be an integer. Exam tip: while finding extrema of a sequence, use only valid integer indices.

The required condition is \(9-2n<-15\). This gives \(-2n<-24\), so \(n>12\). Since \(n\) must be a positive integer, the smallest value greater than 12 is 13. Therefore, the 13th term is the first term less than \(-15\). The 12th term equals \(-15\), so it is not correct. Exam tip: for “less than,” retain the strict inequality \(<\).

The successive differences are \(3,5,7,9\), which are consecutive odd numbers. The successive differences of \(n^2\) also follow \(3,5,7,\ldots\), so the term has the form \(n^2+c\). Using the first term, \(1^2+c=5\), giving \(c=4\). Hence, the \(n\)th term is \(n^2+4\). The option \(n^2+3\) gives 4 as the first term, so it is incorrect. Exam tip: for a sequence involving squares, check the first and second differences.

Given \(a_n=rn+s\), we have \(a_{11}-a_6=5r\). Thus, \(57-32=25=5r\), so \(r=5\). Using \(a_6=6(5)+s=32\), we get \(s=2\). Hence, \(a_{20}=20(5)+2=102\). Option 100 ignores the constant term \(s\). Exam tip: subtracting two terms eliminates \(s\) and helps find \(r\) quickly.

Given \(a_n=n^3\), we get \(a_5=5^3=125\) and \(a_3=3^3=27\). Therefore, \(a_5-a_3=125-27=98\). Option 96 may result from an error in evaluating the cubes or in subtraction. Exam tip: find each term separately before taking their difference.

Given \(a_n=4n^2+n\), we have \(a_{n+1}=4(n+1)^2+(n+1)\). Hence, \(a_{n+1}-a_n=4[(n+1)^2-n^2]+1=4(2n+1)+1=8n+5\). Therefore, \(8n+5\) is correct. A result such as \(8n+3\) can arise from an error in expanding the constant terms. Exam tip: while finding \(a_{n+1}\), replace every occurrence of \(n\) with \(n+1\).

These terms are cumulative sums of squares (1^2), (1^2+2^2), (1^2+2^2+3^2). In cumulative patterns, inspect the added parts.

The given terms are \(3^2,4^2,5^2,6^2,7^2\), respectively. For term number \(n=1\), the base of the square is \(3\), so the base follows the pattern \(n+2\). Hence, the nth term is \((n+2)^2\). For example, at \(n=2\), it gives \((2+2)^2=16\). The expression \(n^2+8\) gives 12 for the second term, so it is not correct. Exam tip: write the initial terms as perfect squares and observe the sequence of their square roots.

The governing concept is evaluation of an explicit sequence rule at a specified index. The notation a₄ means that n must be replaced by 4 throughout the formula, not only in one part. Thus a₄ = (4² + 1)/4 = (16 + 1)/4 = 17/4. The fraction is already in simplest form because 17 is prime and is not divisible by 4. Option B would result from incorrectly using 4² − 1, while option C changes the numerator without justification. Option D, 9/2, is equal to 18/4, so it also does not follow from the substitution. Therefore option A is correct.

Putting n = 5, a_5 = 2(5) + (-1)^5 = 10 - 1 = 9. Since 5 is odd, (-1)^5 = -1. The value 11 would result from incorrectly taking (-1)^5 as +1. Exam tip: remember that (-1)^n is +1 for even n and -1 for odd n.

To find the position of 55, put \(a_n=55\): \(6n-11=55\). Thus, \(6n=66\), so \(n=11\). Therefore, 55 is the 11th term of the sequence. For example, putting \(n=10\) gives \(49\), not 55. Exam tip: To find the term number of a given value, equate \(a_n\) to that value and solve for \(n\).

The governing concept is identifying an explicit rule by testing the index values. For option C, put n = 1: 1² + 3(1) = 4; n = 2 gives 4 + 6 = 10; n = 3 gives 9 + 9 = 18; n = 4 gives 16 + 12 = 28; and n = 5 gives 25 + 15 = 40. Thus every displayed term is reproduced by aₙ = n² + 3n. The other options fail immediately: option A gives 3 for n = 1, option B gives 4 but then 10 for n = 2? Actually it gives 10 at n = 2 but fails at n = 3, and option D gives 5 for n = 1. Hence option C is the only rule matching the sequence.