nth term

The governing concept is the general term of an arithmetic progression. The first term is a=5 and the common difference is d=15−5=10. For an arithmetic progression, the nth term is aₙ = a+(n−1)d. Substituting the values gives aₙ = 5+(n−1)10 = 5+10n−10 = 10n−5. Checking n=1 gives 5, n=2 gives 15, n=3 gives 25, and n=4 gives 35, so the formula matches every displayed term. Hence option A is correct. Option B begins with 15 rather than 5, option C begins with 15 and has the wrong constant term, and option D gives −5 for n=1; these formulas do not generate the stated sequence.

To find the seventh term, substitute \(n=7\): \(a_7=\frac{3(7)+1}{2}=\frac{22}{2}=11\). Therefore, the correct answer is 11. Option 10 could result from omitting the addition of 1 in the numerator. Exam tip: after substituting the value of \(n\), use brackets and follow the order of operations carefully.

The sequence is not arithmetic because its differences are 6, 10, and 14, which are not constant. Instead, compare each term with the square of its position: the first term is 2 times \(1^2\), the second is 2 times \(2^2\), the third is 2 times \(3^2\), and the fourth is 2 times \(4^2\). This reveals the general pattern.

Thus the nth term is \(a_n=2n^2\). Checking the formula gives \(2(1)^2=2\), \(2(2)^2=8\), \(2(3)^2=18\), and \(2(4)^2=32\), exactly matching the given terms. Therefore, option A is correct. The linear choices cannot produce the increasing changes seen in this sequence.

For the ninth term, substitute 9 for n in the given rule: \(a_9=4\times 9+7=36+7=43\). Therefore, 43 is the correct answer. The value 41 would result from an error while adding 7. Exam tip: after substituting the term number in an nth-term formula, perform multiplication before addition.

For \(a_n=3n+1\), \(a_{n+1}-a_n=[3(n+1)+1]-(3n+1)=3\), a constant. Hence it is an AP. For \(n^2+1\), the differences change. Exam tip: a linear form \(pn+q\) represents an AP.

In this geometric progression, the first term is \(a=3\) and the common ratio is \(r=9/3=3\). Using \(a_n=ar^{n-1}\), \(a_5=3\times 3^{4}=243\). Here, 81 is the fourth term, while 729 is the sixth term. Exam tip: for the \(n\)th term, the exponent of the common ratio is always \(n-1\).

Observe the terms: \(3=1^2+2\), \(6=2^2+2\), \(11=3^2+2\), and \(18=4^2+2\). Hence, the \(n\)th term is \(a_n=n^2+2\). The expression \(2n+1\) would give a constant difference of \(2\), but this sequence has differences \(3,5,7,\ldots\). Exam tip: sequences based on squares have consecutive differences that are odd numbers.

(a_n=12+(n-1)5=5n+7), so (5n+7=57) gives (n=10). Set the rule equal to the given term to find its position.

This is an arithmetic progression because the difference between consecutive terms is \(7\). Here, \(a=2\), \(d=7\), and \(n=14\). Therefore, \(a_{14}=a+(n-1)d=2+(14-1)\times7=2+91=93\). Option 91 uses only \(13\times7\) and does not add the first term, 2. Exam tip: For the \(n\)th term, use \(n-1\) common differences, not \(n\).

The first term is (4) and the ratio is (3), so (a_n=4\cdot3^{n-1}). In the geometric rule, the exponent is (n-1).

Substitute 8 for n: a_8=8^2+2(8)=64+16=80. Therefore, 80 is the correct answer. The value 84 may result from an addition error. Exam tip: When finding a term, substitute the value of n everywhere in the formula before evaluating powers and multiplication.

Given (a_n=40-3n), substitute n=12 to find the 12th term: (a_{12}=40-3(12)=40-36=4). Hence, the correct answer is 4. The value 6 can result from an incorrect multiplication or subtraction. In exams, multiply first and then subtract.

The terms are \(2^2,3^2,4^2,5^2\), respectively. For \(n=1\), the first term is \((1+1)^2=4\); hence the \(n\)th term is \((n+1)^2\). Although \(n^2+3\) gives 4 for the first term, it gives 7 when \(n=2\), not the required second term 9. Exam tip: verify a proposed formula using at least the first two terms.

Given \(a_n=8n+3\), \(a_7=8(7)+3=59\) and \(a_2=8(2)+3=19\). Therefore, \(a_7-a_2=59-19=40\). Option 36 may result from an error in multiplication or subtraction. Exam tip: for questions involving two terms, substitute each value of \(n\) separately before finding their difference.

In this geometric progression, the first term is \(a=2\) and the common ratio is \(r=8/2=4\). The \(n\)th term is \(a_n=a r^{n-1}\). Therefore, \(a_5=2\times4^{5-1}=2\times4^4=512\). Choosing 256 results from using one power too few. Exam tip: in the \(n\)th-term formula, the exponent of the common ratio is always \(n-1\).

This is an arithmetic progression with first term \(a=3\) and common difference \(d=7\). Its \(n\)th term is \(a_n=a+(n-1)d=3+7(n-1)=7n-4\). Setting \(7n-4=80\) gives \(7n=84\), so \(n=12\). Hence, 80 is the 12th term. The 11th term is \(73\), so it is not correct despite being a close option. Exam tip: When the position of a given term is asked, equate \(a_n\) to that term and solve for \(n\).

Given \(a_n=4^n\), substitute \(n=3\): \(a_3=4^3=4\times4\times4=64\). Option 16 is the value of \(4^2\), not \(4^3\). Exam tip: For an nth-term question, first substitute the required term number for \(n\), then evaluate the power.

The successive differences are \(4,6,8\), increasing by \(2\) each time, so the rule should be quadratic. Substituting \(n=1,2,3,4\) in \(n^2+n+4\) gives \(6,10,16,24\), respectively. In contrast, \(n^2+5\) gives \(9\) as the second term, so it does not fit. Exam tip: when second differences are constant, test a rule containing \(n^2\).

Given \(a_n=9n-2\), \(a_5=9\times5-2=43\) and \(a_3=9\times3-2=25\). Therefore, \(a_5+a_3=43+25=68\). Option 66 can result from an error while subtracting in one of the terms. Exam tip: Substitute the value of \(n\) separately for each required term before adding.

This is an arithmetic progression with first term \(a=13\) and common difference \(d=19-13=6\). Its \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{18}=13+(18-1)\times6=13+102=115\). Option 109 uses 16 common differences, so it is the 17th term. Exam tip: Remember to use \(n-1\), not \(n\), in the nth-term formula.