Geometric Progression

In a geometric progression, the common ratio is the number by which each term is multiplied to obtain the next term. Compute the quotient of consecutive terms: 18/6=3. Checking further gives 54/18=3 and 162/54=3, so the same multiplier is used throughout the sequence. Therefore the common ratio is r=3.

Option C is consequently correct. The ratio is not found by subtracting terms; the differences are 12, 36, and 108, which are not constant. Nor should the first term itself be chosen as the ratio. Dividing any nonzero term by the preceding term gives the same result here, confirming that the sequence is geometric with positive ratio 3.

Each term is multiplied by (4) so the next term is (256\times4=1024). In exams first identify the common ratio.

The defining property of a geometric progression is that each term after the first is obtained by multiplying the preceding term by the same common ratio. Here the first term is a₁ = 7 and the common ratio is r = 4. Hence the second term is a₂ = a₁r = 7 × 4 = 28, so option C is correct. Adding 4 to 7 gives 11, but addition describes an arithmetic progression rather than a geometric one. The value 21 would require a ratio of 3, not 4, and 32 is not produced by applying the stated rule to the first term. Checking the ratio 28/7 = 4 confirms that 28 satisfies both conditions exactly.

The ratio of consecutive terms is (81\div243=\frac{1}{3}). In exams a decreasing geometric progression can have a fractional ratio.

A sequence is a geometric progression when the ratio of every term to the immediately preceding term is constant. Check the consecutive ratios here: 25/5 = 5, 125/25 = 5, and 625/125 = 5. Since all these ratios are equal, the sequence is a geometric progression with common ratio 5. Therefore, option A is correct. The number 25 is a term of the sequence, not the common ratio, so option B confuses a term with a multiplier. Option C is contradicted by the equal ratios, and option D is false because the terms increase from 5 to 625 rather than decrease. The defining test is equal consecutive ratios.

In a geometric progression, the ratio of each term to the preceding term must remain constant. Here, \(6\div2=3\), but \(12\div6=2\) and \(20\div12=\frac{5}{3}\). Since these ratios differ, the sequence is not a geometric progression. Increasing terms or even terms do not make a sequence a GP. Exam tip: calculate ratios of consecutive terms and check whether they are equal.

The governing concept is the identification of the first term of a geometric progression. In a GP, the first term is denoted by a, and every later term is obtained by multiplying the preceding term by a fixed common ratio r. The displayed order is important: the first number written is the first term. Here the sequence starts with 12, so a = 12. The common ratio confirms the pattern because 36 ÷ 12 = 3, 108 ÷ 36 = 3, and 324 ÷ 108 = 3. Thus 3 is the common ratio, not the first term; 36 and 108 are later terms. Therefore, option B is correct.

The \(n\)th term of a geometric progression is \(T_n=ar^{n-1}\). Therefore, \(T_3=ar^2=2\times6^2=2\times36=72\). Here, \(36\) is only the value of \(r^2\); it must be multiplied by the first term, \(a=2\). Exam tip: for the third term, use \(ar^2\) directly.

Each term is multiplied by (\frac{1}{2}) so the next term is (16\times\frac{1}{2}=8). In exams identify fractional ratios in decreasing sequences.

The common ratio is (3) so the fifth term is (108\times3=324). In exams multiply the last known term by the ratio.

The ratio of (54) and (162) is (3) so the missing term is (54\div3=18). In exams apply the ratio backward too.

The first term is (5) and the ratio is (4) so (a_n=5\cdot4^{n-1}). In exams use (a_n=ar^{n-1}).

The nth term is \(a_n=6^n\). Substituting \(n=2\), we get \(a_2=6^2=36\). Note that \(216=6^3\), so it is \(a_3\), not \(a_2\). Exam tip: first substitute the required term number for \(n\) before evaluating the power.

The first term is (8) and the ratio is (5) so (a_n=8\cdot5^{n-1}). In exams use (a_n=ar^{n-1}).

In a geometric progression, the next term is found by multiplying the current term by the common ratio. Thus the governing formula for the second term is a₂ = a₁r. Substituting a₁ = 54 and r = 1/3 gives a₂ = 54 × 1/3 = 18. Therefore option B is correct. Because the ratio is less than 1, the sequence decreases from 54 to 18 rather than increasing. Option A would involve dividing 54 by 6, which is not the given rule. Option C would result from multiplying by 1/2, and option D incorrectly uses the reciprocal ratio 3. The calculation and the expected decrease agree.

The ratios (5\div2=\frac{5}{2}) and (10\div5=2) are not equal. In exams equal ratio is necessary for a geometric progression.

The common ratio of a geometric progression is the quotient obtained by dividing any term by the immediately preceding term. Using the first two terms, r = 45 ÷ 15 = 3. The result is verified by the later pairs: 135 ÷ 45 = 3 and 405 ÷ 135 = 3. Since the quotient remains constant, the sequence is indeed geometric and its common ratio is 3. Therefore option B is correct. If r were 2, the second term after 15 would be 30; if r were 5, it would be 75. Option D, 15, is the first term rather than the multiplier. Checking several consecutive pairs prevents confusing a term with the ratio.

In a geometric progression, the ratio of each term to the preceding term remains constant. In option A, \(6/3=12/6=24/12=2\), so it is a GP. Option B has a constant difference, not a ratio. Exam tip: compare ratios of consecutive terms.

The common ratio is (\frac{1}{2}) so the fifth term is (18\times\frac{1}{2}=9). In exams simplify fractions.

The terms are (5,15,45,135,405) so (405) is the fifth term. In exams list small terms in order to check.