nth term

Write the terms as squares: \(4=2^2\), \(9=3^2\), \(16=4^2\), and \(25=5^2\). Since the first term is \(2^2\), the \(n\)th term is \((n+1)^2\). Using \(n^2\) would give the first term as \(1\), which does not match the sequence. Exam tip: always test a proposed nth-term formula with \(n=1\).

The governing concept is evaluating an explicit sequence formula at two indices and subtracting the results. First calculate a₈: 3(8²) − 2(8) + 1 = 3 × 64 − 16 + 1 = 192 − 16 + 1 = 177. Next calculate a₅: 3(5²) − 2(5) + 1 = 3 × 25 − 10 + 1 = 75 − 10 + 1 = 66. Hence a₈ − a₅ = 177 − 66 = 111. Therefore option D is correct. Computing both terms separately avoids confusing the index with the value. The other choices do not result from the correct substitution and subtraction.

Check the terms: \(7=1^2+6\), \(10=2^2+6\), \(15=3^2+6\), \(22=4^2+6\), and \(31=5^2+6\). Therefore, the \(n\)th term is \(n^2+6\). The expression \(2n+5\) is linear, so it would have a constant first difference, whereas this sequence has differences \(3,5,7,9\). Exam tip: If first differences are consecutive odd numbers, test a rule involving \(n^2\).

The increase over (5) positions is (25), so the common difference is (5). Then (a_4=18) gives the constant term (-2).

Given \(2n^2+3n=65\), we get \(2n^2+3n-65=0\). Factoring gives \((n-5)(2n+13)=0\), so \(n=5\) or \(n=-\frac{13}{2}\). Since a term number in a sequence must be a positive integer, \(n=5\) is correct. For example, \(n=6\) gives 90, not 65. Exam tip: after solving, always check whether the value is a valid positive integer term number.

Observe the numerator and denominator patterns separately. The numerators are 1, 2, 3, 4, so the numerator of the nth term is n. The denominators are 2, 3, 4, 5, each one greater than its corresponding numerator, so the denominator is n + 1. Therefore the nth term is n/(n + 1), making option A correct.

The governing concept is an explicit rule for the nth term. The absolute values of the terms are 1, 4, 9, 16, and 25, which are 1², 2², 3², 4², and 5². The signs begin negative and then alternate: negative for odd n and positive for even n. Since (−1)ⁿ is negative when n is odd and positive when n is even, the required rule is aₙ = (−1)ⁿn². Checking gives a₁ = −1, a₂ = 4, a₃ = −9, and a₄ = 16. Option A gives the opposite sign, option B misses all signs, and option D does not reproduce the listed terms.

For a term to be negative, \(12-3n<0\). Thus \(3n>12\), so \(n>4\). The smallest natural value of \(n\) is 5; hence \(a_5=12-3(5)=-3\) is the first negative term. Note that \(a_4=0\), which is not negative. Exam tip: For the “first” term, choose the smallest positive integer satisfying the inequality.

Given \(a_n=pn+q\), we have \(a_7-a_3=4p=27-11=16\), so \(p=4\). Using \(a_3=11\), \(12+q=11\), hence \(q=-1\). Therefore, \(a_{12}=4\times12-1=47\). Choosing 45 would correspond to \(4n-3\), which does not satisfy \(a_3=11\). Exam tip: when two terms are given, subtract them first to find \(p\).

The nth term is \(a_n=2^n+1\). Substituting \(n=6\), we get \(a_6=2^6+1=64+1=65\). Option 63 would result from \(2^6-1\), but the given expression requires adding 1. Exam tip: evaluate the exponent first, then perform addition or subtraction.

The successive differences are 6, 8, 10, and 12, so the second difference is constantly 2. This indicates a quadratic expression with leading coefficient 1, so write the term as \(a_n=n^2+bn+c\). Use the first two terms. For n=1, \(1+b+c=5\), giving \(b+c=4\). For n=2, \(4+2b+c=11\), giving \(2b+c=7\). Subtracting gives \(b=3\), and then \(c=1\).

Thus \(a_n=n^2+3n+1\), which is option B. A quick check gives 5 for n=1, 11 for n=2, and 19 for n=3. The constant second difference confirms the quadratic form, while direct substitution determines the remaining coefficients. The supplied answer and explanation are accurate.

To find the position of 79, put a_n=79. Then 4n-1=79, so 4n=80 and n=20. Hence, 79 is the 20th term. The 19th term is 4(19)-1=75, so it is not correct. Exam tip: To find the term number of a given value, equate the general term to that value and solve for n.

The consecutive differences are \(2,4,6,8,\ldots\), so the added differences up to the \(n\)th term are \(2(k-1)\). Hence, \(a_n=1+2(1+2+\cdots +(n-1))=1+n(n-1)=n^2-n+1\). The close distractor \(n^2+n-1\) gives \(5\) when \(n=2\), whereas the second term is \(3\). Exam tip: if first differences increase by a constant amount, the general term is usually quadratic in \(n\).

Given \(a_n=n^2+4n\), \(a_{10}=10^2+4(10)=100+40=140\) and \(a_1=1^2+4(1)=1+4=5\). Therefore, \(a_{10}+a_1=140+5=145\). Note that \(140\) is only \(a_{10}\), not the required sum. Exam tip: substitute each value of \(n\) into the formula separately before adding the terms.

Given \(a_n=7n+2\), set the term equal to \(100\): \(7n+2=100\). Thus, \(7n=98\) and \(n=14\). Hence, 100 is the 14th term. For comparison, \(a_{13}=7\times13+2=93\), so the 13th term is not correct. Exam tip: equate the given term value to \(a_n\) and solve for \(n\).

This is an arithmetic sequence because each successive term decreases by 3. Hence, \(a_1=10\) and \(d=-3\). Using \(a_n=a_1+(n-1)d\), we get \(a_n=10+(n-1)(-3)=13-3n\). In option A, putting \(n=1\) gives 7, not the first term 10. Exam tip: verify an nth-term formula by substituting \(n=1\) and checking the first term.

Substitute \(n=7\) in \(a_n=3\cdot2^{n-1}\): \(a_7=3\cdot2^{7-1}=3\cdot2^6=3\cdot64=192\). Therefore, 192 is the correct option. The value 96 results from incorrectly using the exponent 5 instead of 6. Exam tip: simplify the exponent \(n-1\) before evaluating the power.

For a zero term, n(n-2)=0. Hence, n=0 or n=2. Term numbering in a sequence normally starts from 1, so n=0 is not considered a term number. At n=2, a_2=2(2-2)=0; therefore, the second term is the first zero term. Exam tip: Always check the starting value of n for term numbers.

Given (a_n=2n^2-kn), substitute n=4: (a_4=2(4)^2-4k=32-4k). Since (a_4=24), we get (32-4k=24), so (4k=8) and hence (k=2). Therefore, option C is correct. If k=3, then (a_4=20), not 24. Exam tip: When a particular term is given, substitute that term number for n first.

The sequence is \\(6,11,18,27,38,\\ldots\\). Its successive differences are 5, 7, 9, and 11, so the difference increases by 2 each time. This suggests a quadratic expression. Test the supplied option \\(a_n=n^2+2n+3\\): for \\(n=1\\), it gives \\(1+2+3=6\\); for \\(n=2\\), it gives \\(4+4+3=11\\); and for \\(n=3\\), it gives \\(9+6+3=18\\). It also gives 27 and 38 for n = 4 and 5. Thus option B is correct.

The formula can be confirmed through the second differences. For a quadratic \\(an^2+bn+c\\), a constant second difference is expected; here it is 2, so the coefficient of \\(n^2\\) is 1. Using the terms then gives the expression \\(n^2+2n+3\\). The nearby choices fail even at the first or second term. Therefore the supplied answer B and its explanation correctly identify the nth term.