nth term

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 common ratio is (3), so (a_4=6\cdot3^3=162). In a geometric term, the exponent of the ratio is (n-1).

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.

This is an arithmetic progression because each term decreases by 4. Thus, the first term is \(a=30\) and the common difference is \(d=-4\). \(a_9=a+(9-1)d=30+8(-4)=30-32=-2\). Therefore, \(-2\) is the correct answer. The value \(0\) would result from subtracting the common difference only 7 times. Exam tip: for the \(n\)th term, use \(n-1\) common differences.

Given \(a_n=3n+14\) and \(a_n=56\), we get \(3n+14=56\). Subtracting 14 from both sides gives \(3n=42\), so \(n=14\). Hence, 56 is the 14th term of the sequence. If \(n=15\), the term would be \(3\times15+14=59\), not 56. Exam tip: To find the term number, substitute the given value of \(a_n\) into the formula and solve the resulting equation for \(n\).

The common ratio of this GP is \(r=\frac{48}{96}=\frac{1}{2}\). Therefore, the sixth term is \(a_6=96\left(\frac{1}{2}\right)^5=3\). The terms are \(96,48,24,12,6,3\), so 3 is correct. Option 6 is the fifth term, not the sixth. Exam tip: In \(a_n=a r^{n-1}\), use the exponent \(n-1\).

This is an arithmetic progression with first term \(a=9\) and common difference \(d=9\). Hence, \(a_n=a+(n-1)d=9+(n-1)\times9=9n\). The expression \(n+9\) gives 11 when \(n=2\), but the second term is 18. Exam tip: use \(a_n=a+(n-1)d\) for an arithmetic progression.

Given \(a_n=12-5n\), substitute \(n=4\) to find the fourth term: \(a_4=12-5(4)=12-20=-8\). Therefore, \(-8\) is correct. An option such as \(-6\) is incorrect because \(12-20=-8\), not \(-6\). Exam tip: after substituting the value of \(n\), do the multiplication before subtraction.

The consecutive differences are \(7,9,11\), increasing by \(2\) each time, so the sequence follows a quadratic pattern. Substituting \(n=1,2,3,4\) in \(a_n=n^2+4n\) gives \(5,12,21,32\), respectively. \(n^2+4\) matches only the first term and gives \(8\) for the second term. Exam tip: when first differences increase uniformly, test a quadratic nth-term rule.