nth term

To find the 20th term, substitute n=20 in the given rule: \(a_{20}=5\times20-7=100-7=93\). Therefore, the correct answer is 93. The value 95 would result if 5 were subtracted, but the rule requires subtracting 7. Exam tip: multiply first, then subtract the constant 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 \(n=1\).

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.

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.

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.