Sequences

The rule is: next term = \(3\times\) previous term \(+1\). Therefore, the term after \(121\) is \(3\times121+1=363+1=364\). Option 363 is only three times 121 and misses the added 1. Exam tip: apply the stated rule directly to the last given term.

In \((7,7,7,\ldots)\), the common difference is \(0\) and the common ratio is \(1\), so it is both an AP and a GP. Option B has a fixed ratio but no fixed difference. Exam tip: check both conditions separately.

The consecutive differences are \(10-7=3\), \(16-10=6\), \(28-16=12\), and \(52-28=24\). Each difference is twice the preceding one, so the next difference is \(48\). Therefore, the next term is \(52+48=100\). Option 96 would require a next difference of 44, which does not follow the doubling rule. Exam tip: For such sequences, first list consecutive differences and check their pattern.

It adds consecutive numbers starting from (1), so (a_n=1+\frac{n(n+1)}{2}). Thus (a_{20}=1+\frac{20\times21}{2}=211).

The answer is option C, the 12th term. The sequence 6, 13, 20, 27,... is arithmetic because each term increases by 7. Thus the first term is a=6 and the common difference is d=7. The nth-term formula is a_n=a+(n-1)d. Substituting the target value 83 gives 83=6+(n-1)7. Subtracting 6 gives 77=7(n-1). Dividing by 7 gives n-1=11, so n=12. Option A, 10th, would give 6+9(7)=69, not 83. Option B, 11th, would give 6+10(7)=76. Option C, 12th, gives 6+11(7)=83, so it is correct. Option D, 13th, gives 6+12(7)=90. Therefore C is the only correct option. A useful check is to count the jumps: from 6 to 83 there are 77 units, and 77 contains eleven jumps of 7, so 83 is the twelfth term.

The direct answer is option D: -128. The sequence is geometric because every term is obtained by multiplying the preceding term by -2: 1×(-2)=-2, (-2)×(-2)=4, 4×(-2)=-8, and so on. The first term is a=1 and the common ratio is r=-2. For a geometric sequence, the nth term is \\(a_n=ar^{n-1}\\). Therefore, the eighth term is \\(a_8=1(-2)^7\\). Since the exponent 7 is odd, the result is negative, and \\((-2)^7=-128\\). Option A, 64, has the wrong sign and is not the eighth term. Option B, -64, has the correct negative sign but the magnitude is too small; it corresponds to a different power. Option C, 128, has the wrong positive sign because an odd power of -2 is negative. Option D, -128, has both the correct magnitude and sign, so it is correct. Another safe method is to continue counting: sixth term is 32, seventh is -64, and eighth is 128? Careful: starting from the listed terms, the sixth term is 32, the seventh is -64, and the eighth is 128. Thus the supplied answer D is inconsistent: the actual eighth term is 128, option C. The question should be revised or the answer mapping corrected.

The direct answer is option C: 320. This is a geometric sequence. Each term is multiplied by -2: 5×(-2)=-10, (-10)×(-2)=20, and 20×(-2)=-40. Thus the first term is a=5 and common ratio is r=-2. The nth-term formula is \\(a_n=ar^{n-1}\\). For n=7, \\(a_7=5(-2)^6\\). Now \\((-2)^6=64\\) because an even power has a positive result. Hence \\(a_7=5×64=320\\). Option A, 160, is too small and does not result from the sixth power calculation. Option B, -160, has both an incorrect magnitude and sign. Option C, 320, follows exactly from the formula and is correct. Option D, 640, is twice the correct value and may result from using an incorrect power or multiplying by 10 instead of 5. A direct count confirms it: fifth term is 80, sixth is -160, and seventh is 320. The important exam point is that the negative ratio changes signs alternately, while an even exponent makes the final sign positive.

The next terms are (13,21,34,55), so the (10)th term is (55). Exam tip: write positions when extending Fibonacci-type sequences.

In \(7,7,7,7,\ldots\), each consecutive difference is \(7-7=0\), so it is an AP. Each consecutive ratio is \(7/7=1\), so it is also a GP. \(2,5,8,\ldots\) is only an AP. Exam tip: for a constant sequence, check difference 0 and ratio 1.

Given \(a_n=3n^2\), \(a_8=3\times 8^2=192\) and \(a_6=3\times 6^2=108\). Hence, \(a_8-a_6=192-108=84\). Option 78 is incorrect because the values must be found by squaring the term numbers first. Exam tip: substitute each required value of \(n\) into the general term before subtracting.

The differences between consecutive terms are 10−4=6, 18−10=8, 28−18=10, and 40−28=12. These differences increase by 2 each time, so the next difference is 14. Hence, the next term is 40+14=54. Choosing 52 would give a difference of only 12 and would break the pattern of increasing differences. Exam tip: Write the consecutive differences first to identify the pattern in a sequence.

The next difference is (23), so (56+23=79). Exam tip: identifying the difference of differences is the fastest method here.

Use the stated successive differences in order. Starting from 171, the next term is obtained by adding -22, so x = 171 - 22 = 149. Checking the final step gives 149 + (-27) = 122, which agrees with the last given term. Therefore option A is correct; the other options fail this verification or mishandle the negative differences.

The governing concept is using successive differences to extend a sequence. Subtracting consecutive terms gives 6 − 4 = 2, 9 − 6 = 3, 14 − 9 = 5, 21 − 14 = 7, and 32 − 21 = 11. These are consecutive prime numbers. The next prime after 11 is 13, so the next term must be the current final term plus 13: 32 + 13 = 45. Therefore option C is correct. The distractor 41 would add 9, 43 would add 11 by repeating the previous difference, and 47 would add 15; none follows the stated prime-difference pattern. Writing the differences separately avoids confusing terms with increments.

The governing idea is to separate an interleaved sequence into its odd-position and even-position terms. The odd-position terms are 1, 2, 4, 8, ..., and each doubles. The even-position terms are 5, 10, 20, 40, 80, ..., and each also doubles. Because position 10 is even, we use the even-position subsequence. Its terms are: position 2 gives 5, position 4 gives 10, position 6 gives 20, position 8 gives 40, and position 10 gives 80. Therefore the tenth term is 80, so option D is correct. The values 16 or 32 come from following the odd-position pattern or confusing the original position with the index within one subsequence. The position parity must be checked first.

The governing concept is identifying the continuation rule before finding a requested sum. The successive differences of the given terms are 4, 6, 8, and 10. These differences increase by 2, so the next difference is 12. Therefore the sixth term is 31 + 12 = 43. Now add the first six terms: 3 + 7 + 13 + 21 + 31 + 43. Step by step, 3+7=10, 10+13=23, 23+21=44, 44+31=75, and 75+43=118. Thus option B is correct. This is not an arithmetic progression because the first differences are not constant. An answer such as 112 may result from using an incorrect sixth term, while 114 or 116 can result from incomplete or inaccurate addition.

The governing idea is identifying the pattern and then adding exactly the requested number of terms. The successive differences are 4, 6, 8, 10, so the next difference is 12 and the sixth term is 31 + 12 = 43. Hence the sum of the first six terms is 3 + 7 + 13 + 21 + 31 + 43 = 118. Thus option D is correct; smaller choices omit or miscalculate a term.