Geometric Progression

The key distinction is between an arithmetic pattern with a constant difference and a geometric pattern with a constant ratio. In option B, each term is obtained by multiplying the preceding term by 2: 4 × 2 = 8, 8 × 2 = 16, and 16 × 2 = 32. Thus it follows a multiplication rule. Option A follows addition of 3 at every step, option C follows subtraction of 3, and option D follows addition of 4. These three sequences have constant differences rather than a repeated multiplier. Therefore option B is the only sequence formed by multiplication. The fact that its terms double each time, rather than increasing by a fixed amount, identifies it as a geometric progression.

The governing concept is a geometric progression, because each term is obtained by multiplying the preceding term by the same non-zero constant. The common ratio is r=12/4=3; this is confirmed by 36/12=3 and 108/36=3. For a geometric progression, the nth term is a_n=a_1r^(n−1). Therefore a_7=4×3^6=4×729=2916. Continuing the sequence also gives a_5=324, a_6=972, and a_7=2916. Hence option C is correct. Option A is the sixth term, while options B and D do not result from repeatedly multiplying by the fixed ratio 3. The initial term and ratio are both used, so no addition-based rule is appropriate.

The governing concept is a geometric progression, because the ratio of every term to the preceding term is constant. Here the common ratio is \(r=\frac13\). Starting from the first term, divide by 3 repeatedly: the first four terms are 81, 27, 9, and 3; the fifth term is \(3\div3=1\), and the sixth term is \(1\div3=\frac13\). Using the formula gives the same result: \(a_6=81(\frac13)^5=\frac{81}{243}=\frac13\). Therefore option B is correct. Option A is only the fifth term, option C applies the division one extra time, and option D is impossible because repeated division of a positive number by 3 never produces exactly zero.

The governing concept is a geometric progression, because every term is obtained by multiplying the previous term by the same ratio, 2. With first term 3 and common ratio 2, the nth term is a_n = 3×2^(n−1). To find the first term greater than 500, examine the boundary terms. The eighth term is a_8 = 3×2^7 = 3×128 = 384, which is not greater than 500. The ninth term is a_9 = 3×2^8 = 3×256 = 768, which is greater than 500. Since the progression is increasing, a_9 is the first term satisfying the condition. Therefore option B is correct. Checking only 768 would show that it is large enough, but comparing it with a_8 proves that it is the first such term.

The governing concept is the nth-term formula for a geometric progression. The first term is 5 and the common ratio is 2, so a_n = 5×2^(n−1). The exponent is n−1 because the first term has undergone zero doublings, the second has undergone one doubling, and so forth. To locate 5120, divide by the first term: 5120 ÷ 5 = 1024. Since 1024 = 2^10, we have 5120 = 5×2^10. Comparing this with 5×2^(n−1), n−1 = 10, so n = 11. Therefore option B is correct. Choosing the 10th term incorrectly uses n instead of n−1; choosing 12th or 13th counts extra doublings.

The governing concept is that consecutive terms of a geometric progression have a constant quotient. Divide the second term by the first: \(r=\frac{6}{3}=2\). The remaining terms confirm this, because \(\frac{12}{6}=2\) and \(\frac{24}{12}=2\). Thus option A is correct. A ratio of 3 would produce 3, 9, 27, and so on, while ratios 4 and 6 would produce much larger successive terms. Although the terms visibly double, the formal method is to divide one term by the preceding term and verify that the quotient remains constant.

The \(n\)th term of a GP is \(a_n=ar^{n-1}\). Here, \(a=5\), \(r=3\), and \(n=4\), so \(a_4=5\times3^{4-1}=5\times27=135\). Therefore, 135 is correct. A value such as 90 results from using the exponent or term position incorrectly. Exam tip: to reach the fourth term, multiply the common ratio \(r\) three times.

In (4,12,36,108,\ldots), each term is multiplied by (3). In a geometric progression consecutive ratios are equal.

The common ratio is (\frac{27}{81}=\frac{1}{3}). In a decreasing geometric progression the ratio can be less than (1).

Given \(a_n=2\cdot4^{n-1}\), put \(n=5\): \(a_5=2\cdot4^{5-1}=2\cdot4^4=2\cdot256=512\). Therefore, 512 is correct. The value 256 is only \(4^4\); the initial factor 2 must also be multiplied. Exam tip: substitute the given value of \(n\) into \(n-1\) first.

The direct answer is C: 224 is the sixth term. The sequence starts at 7 and doubles each time: 7, 14, 28, 56, 112, 224. Thus the positions are first through sixth, and 224 is at position 6. Using the formula, \(a_n=7\cdot2^{n-1}\). Since \(224=7\cdot32=7\cdot2^5\), we have \(n-1=5\), so \(n=6\). Option A is wrong because the fourth term is 56. Option B is wrong because the fifth term is 112. Option C is correct because the sixth term is 224. Option D is wrong because the seventh term would be 448. The common mistake is to count the number of doublings as the term number; the first term already exists before any doubling. For small values, writing the terms in order is the quickest reliable check.

In (2,4,8,15,\ldots), the ratios (2,2,\frac{15}{8}) are not equal. A geometric progression must have a constant ratio.

The first term is (6), and multiplying by (2) each time gives (6,12,24,48). In a geometric progression each next term is formed by multiplication.

In a geometric sequence, the common ratio is found by dividing any term by the preceding term. Here, \(r=\frac{-6}{2}=-3\). Checking further, \(\frac{18}{-6}=-3\) and \(\frac{-54}{18}=-3\). Hence, the common ratio is \(-3\). Option \(3\) is a close distractor, but it does not account for the alternating signs. Exam tip: divide consecutive terms and verify the ratio with another pair.

The common ratio is (3), so the next term is (36\cdot3=108). Multiply the previous term by the ratio for the next term.

For the middle term (x), (x^2=3\cdot27=81), so (x=9). For positive terms take the positive geometric mean.

The first four terms are 5, 15, 45, and 135. Therefore, their sum is \(5+15+45+135=200\). Hence, 200 is correct. The value 180 is the sum of only the first three terms, so it is a close but incorrect option. Exam tip: when there are only a few terms, direct addition is usually the quickest and safest method.

In a geometric progression, the second term is \(a_2=a_1r\). Thus, \(10=a_1\times2\), so \(a_1=10/2=5\). Taking \(10\) as the first term is incorrect because it is given as the second term. Exam tip: To find a preceding term, divide the next term by the common ratio.

The general term of a GP is
\(a_n=a_1r^{n-1}\). Thus,
\(a_4=4r^3=108\), so
\(r^3=27\) and
\(r=3\). If the ratio were 2, the fourth term would be
\(4\times2^3=32\), not 108. Exam tip: from
\(a_m\) to
\(a_n\), the exponent of the ratio is
\(n-m\).

This is a geometric sequence because each term is twice the preceding term, so the common ratio is \(r=2\). The terms are \(\frac{1}{2},1,2,4,8,16\); therefore, the 6th term is \(16\). Note that \(8\) is the 5th term, making it a close but incorrect option. Exam tip: Start counting terms from the first given term.

The common ratio is (\frac{1}{2}), so the fifth term is (4). In a decreasing geometric progression multiply by the ratio repeatedly.

The \(n\)th term of a geometric progression is \(a_n=ar^{n-1}\). Thus, \(a_4=3\times5^{4-1}=3\times125=375\). \(125\) is only \(5^3\); it misses multiplication by the first term, \(a=3\). Exam tip: in \(a_n\), the exponent of \(r\) is always \(n-1\), not \(n\).

The direct answer is B: \\(64,16,4,1,\\ldots\\). The common ratio of a geometric progression is found by dividing a term by the immediately preceding term. For A, \\(8/4=2\\), so its ratio is 2, not \\(1/4\\). For B, \\(16/64=1/4\\), and also \\(4/16=1/4\\) and \\(1/4=1/4\\), so the ratio remains \\(1/4\\); B is correct. For C, \\(4/1=4\\), so its ratio is 4. For D, \\(4/8=1/2\\), so its ratio is \\(1/2\\), not \\(1/4\\). The fact that terms decrease is not enough; the exact multiplier must be checked. Memory cue: divide second by first, then third by second; the same result must appear.

In this GP, the first term is 1 and the common ratio is 3. Its terms are 1, 3, 9, 27, 81; therefore, 81 is the fifth term. The fourth term is 27, so it is a close but incorrect option. Exam tip: list the terms in order or recognise \(81=3^4\).

For the third term, substitute \(n=3\): \(a_3=9\left(\frac{1}{3}\right)^{3-1}=9\left(\frac{1}{3}\right)^2=9\times\frac{1}{9}=1\). The option \(3\) results from incorrectly using the exponent as 1. Exam tip: after substituting the term number in \(a_n\), calculate \(n-1\) first.