nth term

Given \(a_n=3\cdot4^{n-1}\), substitute \(n=4\): \(a_4=3\cdot4^{4-1}=3\cdot4^3=3\cdot64=192\). Hence, 192 is correct. The value 96 may result from an incorrect multiplication after evaluating the power. Exam tip: first substitute the term number into \(n-1\), then simplify step by step.

This is an arithmetic progression with first term \(a=14\) and common difference \(d=4\). Therefore, \(a_n=a+(n-1)d=14+4(n-1)=4n+10\). Putting \(a_n=70\), we get \(4n+10=70\), so \(4n=60\) and \(n=15\). For \(n=14\), the term would be \(66\), so it is a close but incorrect option. Exam tip: to find the position of a term, first write \(a_n=a+(n-1)d\).

For the sixth term, substitute n=6: \(a_6=6^2+2(6)-1=36+12-1=47\). Therefore, the correct answer is 47. The value 49 would result from omitting the final \(-1\), so it is incorrect. Exam tip: Substitute the value of n in every occurrence of n and follow the order of operations carefully.

The first term of this GP is 2 and the common ratio is 5. Therefore, its nth term is \(a_n=2\times5^{n-1}\). From \(2\times5^{n-1}=1250\), we get \(5^{n-1}=625=5^4\), so \(n=5\). Hence, 1250 is the fifth term. The fourth term is \(250\), making it a close but incorrect option. Exam tip: To find a position in a GP, first write \(a_n=ar^{n-1}\).

The governing concept is the general term of an arithmetic progression. The first term is a=6, and the common difference is d=16−6=10. Applying aₙ=a+(n−1)d gives aₙ=6+(n−1)10=6+10n−10=10n−4. Verification is straightforward: n=1 gives 6, n=2 gives 16, n=3 gives 26, and n=4 gives 36. Therefore option A is correct. Option B gives 14 at n=1, option C gives 16 at n=1 and changes by 6 rather than 10, and option D gives 11 at n=1 with common difference 1. Each distractor fails either the first-term check, the difference check, or both.

Given \(a_n=\frac{5n-1}{2}\), substitute \(n=7\): \(a_7=\frac{5(7)-1}{2}=\frac{35-1}{2}=\frac{34}{2}=17\). Hence, 17 is correct. The value 16 could result from incorrectly calculating \(35-1\) as 32. Exam tip: In an nth-term formula, substitute the value of \(n\) carefully and simplify the operations in order.

The sequence follows a square-number pattern multiplied by 3. The first term is \(3\times1^2\), the second is \(3\times2^2\), the third is \(3\times3^2\), and the fourth is \(3\times4^2\). The differences are 9, 15, and 21, so this is not an arithmetic sequence with a constant difference; the square pattern is the useful observation.

Therefore, the nth term is \(a_n=3n^2\). Substitution verifies every displayed term: \(3(1)^2=3\), \(3(2)^2=12\), \(3(3)^2=27\), and \(3(4)^2=48\). Hence option A is correct. A formula such as \(3n+9\) is linear and cannot produce these values for all positions, even if it may match one term by chance.

The governing concept is evaluating an explicit formula for selected terms of a sequence. Substitute n=4 into aₙ=11n+2 to obtain a₄=11(4)+2=44+2=46. Then substitute n=6 to obtain a₆=11(6)+2=66+2=68. Adding the two required terms gives a₄+a₆=46+68=114. Therefore option C is correct. A useful check is to combine algebraically: (11×4+2)+(11×6+2)=11(10)+4=110+4=114. Option A may result from an arithmetic error, option B omits the two constant terms, and option D adds too much; none equals the correctly calculated sum.

The terms are (7,14,28,56,112,224,448), so (448) is the seventh term. In a geometric progression, multiply by the same ratio each time.

Substitute n=5: a_5=2(5)^2+3(5)=2×25+15=50+15=65. Therefore, 65 is the correct answer. An answer such as 60 usually results from an error in evaluating the square or multiplication. Exam tip: substitute the value first, then apply powers, multiplication, and addition in order.

The first term is (40) and the difference is (-5), so (a_n=40+(n-1)(-5)=45-5n). In a decreasing sequence, take the difference as negative.

Given \(a_n=\frac{n^2+n}{2}\). Substituting \(n=10\), \(a_{10}=\frac{10^2+10}{2}=\frac{100+10}{2}=\frac{110}{2}=55\). Hence, the correct answer is 55. Getting 50 is a common error caused by evaluating \(10^2\) incorrectly or not adding the numerator correctly. Exam tip: substitute the value of \(n\) first, then evaluate the power, addition, and division in order.

Given \(a_n=5\cdot2^{n-1}\), substitute \(n=6\): \(a_6=5\cdot2^{6-1}=5\cdot2^5=5\cdot32=160\). Hence, option C is correct. \(80\) would result from using exponent \(4\), but here \(n-1=5\). Exam tip: In an nth-term formula, substitute the value of \(n\) first and then evaluate the exponent carefully.

For the fifth term, substitute n=5. Thus, a_5=5+6=11, so 11 is correct. The value 10 would result from using n=4, which gives the fourth term. Exam tip: In an nth-term question, first substitute the requested term number for n.

For \(a_n=4n+1\), putting \(n=1\) gives the first term as 5, and consecutive terms differ by 4. Hence the sequence is 5, 9, 13, 17, .... Option C has common difference 4 but starts at 1. Exam tip: check \(n=1\) first.

This is an arithmetic progression with first term \(a=4\) and common difference \(d=4\). Therefore, \(a_{12}=a+(12-1)d=4+11\times4=48\). Hence, 48 is correct. The value 44 can result from incorrectly adding only 10 differences; from the first term to the twelfth term, there are 11 differences. Exam tip: use \(a_n=a+(n-1)d\) and substitute the values carefully.

Putting \(n=1\) gives the first term \(4(1)-3=1\), and each next term increases by 4. Hence 1, 5, 9, 13, ... is correct. Option B also has common difference 4, but its first term is 4. In exams, check both the first term and the difference.

For the fifth term, substitute \(n=5\) in the formula: \(a_5=5^2+2=25+2=27\). Hence, 27 is correct. The value 25 is only \(5^2\); the additional 2 must also be included. Exam tip: Substitute the required value of \(n\) first, then follow the order of operations carefully.

The governing concept is identifying the rule of a geometric sequence. Each displayed term is twice the preceding term: 2×2=4, 4×2=8, and 8×2=16. Continuing the same rule gives the fifth term as 16×2=32. Equivalently, the terms can be written as 2¹, 2², 2³, 2⁴, so the fifth term is 2⁵=32. Therefore option B is correct. Option A does not follow the repeated doubling pattern, option C is not produced by multiplying 16 by 2, and option D is the sixth term because 32×2=64. This check also shows why simply adding a fixed number would be inappropriate for this sequence.

Substituting \(n=1,2,3\) in \(2n-1\) gives \(1,3,5\), the odd natural numbers. In contrast, \(2n\) generates even numbers. Exam tip: verify a general term by listing its first three terms.