Geometric Progression

The governing concept for a geometric progression is a constant ratio between every pair of consecutive terms. To find that ratio, divide a term by the term immediately before it. Here, 6 divided by 2 equals 3; 18 divided by 6 also equals 3; and 54 divided by 18 again equals 3. Since the quotient remains constant, the common ratio is r = 3. Therefore option B is correct. Option A is the first term, not the ratio. Option C is obtained by confusing multiplication or subtraction with division, and option D does not describe the relationship between any consecutive terms. The sequence is geometric because each term is three times the preceding term.

Each term is multiplied by (2), so the next term is (40\times2=80). In exams, first identify the common ratio.

In a geometric progression, each term after the first is obtained by multiplying the preceding term by the common ratio. Therefore, to find the second term, multiply the first term by the given ratio. This rule is different from an arithmetic progression, where a fixed number is added instead.

Here the first term is \(4\) and the common ratio is \(2\). Hence the second term is \(4\times2=8\). The sequence begins \(4,8,16,32,\ldots\). Thus, option C is correct. The value 6 would come from adding 2, and 10 would come from adding 6; neither follows the geometric-progression rule given here.

The ratio of consecutive terms is (27\div81=\frac{1}{3}). In exams, a decreasing geometric progression can have a fractional ratio.

A sequence is a geometric progression when the ratio of each term to the preceding term is constant. Check the consecutive quotients: 6 divided by 3 equals 2, 12 divided by 6 equals 2, and 24 divided by 12 equals 2. Because every checked ratio is the same, the sequence is geometric and its common ratio is 2. Hence option A is correct. Option B confuses the ratio with the first term or with an unrelated multiplier. Option C is false because the ratios are equal, and option D is false because the terms are increasing, not decreasing. The repeated multiplication by 2 is the defining feature here.

The defining condition for a geometric progression is a constant multiplicative ratio between each pair of consecutive nonzero terms. In this sequence, 4 ÷ 2 = 2, 6 ÷ 4 = 3/2, and 8 ÷ 6 = 4/3. These ratios are different, so the sequence does not have one common ratio and is not geometric. Therefore option C is correct. The sequence is instead an arithmetic progression because its consecutive differences are constant: 4 - 2 = 2, 6 - 4 = 2, and 8 - 6 = 2. Having first term 2, increasing terms, or even terms does not decide whether a sequence is geometric; those are only observations. The decisive test is equality of consecutive ratios, not equality of differences.

The governing concept is identification of the first term in a geometric progression. When a sequence is written from left to right, its first term is the number at the beginning, usually denoted by a. The given progression begins with 7, so its first term is a = 7. The repeated multiplication by 3 is a separate property: 21 ÷ 7 = 3, 63 ÷ 21 = 3, and 189 ÷ 63 = 3, showing that the common ratio is 3. It does not alter the first term. Hence option B is correct. Options C and D are the second and third terms, respectively, while option A is the reciprocal of the common ratio, not a term of the displayed progression.

In a geometric progression, each term is obtained by multiplying the preceding term by the same common ratio r. If the first term is a, the terms begin as a, ar, and then \(ar^2\). Therefore the third term is found using the ratio twice, not once. With a=3 and r=4, the third term is \(ar^2=3\times4^2=3\times16=48\).

The sequence can also be written step by step: the first term is 3, the second is \(3\times4=12\), and the third is \(12\times4=48\). Hence option C is correct. The value 12 is the second term, while 24 does not follow the required multiplication pattern. The value 64 is only \(4^3\), so it ignores the first term 3. The exponent in the formula is one less than the term number.

Each term is multiplied by (\frac{1}{2}), so the next term is (8\times\frac{1}{2}=4). In exams, identify fractional ratios in decreasing sequences.

The common ratio is (3), so the fifth term is (54\times3=162). In exams, multiply the last known term by the ratio.

The ratio of (36) and (108) is (3), so the missing term is (36\div3=12). In exams, apply the ratio backward too.

Here, the first term is \(a=3\) and the common ratio is \(r=3\). Thus, \(a_n=ar^{n-1}=3\times3^{n-1}=3^n\). The option \(a_n=3^{n-1}\) gives 1 as the first term, not 3. Exam tip: substitute \(n=1\) to check whether a proposed general term gives the first term of the sequence.

Given \(a_n=2^n\). For the fifth term, substitute \(n=5\): \(a_5=2^5=32\). Therefore, 32 is correct. The value 16 equals \(2^4\), so it represents the fourth term, not the fifth. Exam tip: To find a particular term of a sequence, substitute its term number directly for \(n\) in the given formula.

The direct answer is B: \(a_n=5\cdot3^{n-1}\). In a geometric progression, every term is obtained by multiplying the previous term by the same common ratio. Here \(15/5=3\), \(45/15=3\), and \(135/45=3\), so the first term is \(a=5\) and the ratio is \(r=3\). The general formula is \(a_n=ar^{n-1}\), hence \(a_n=5\cdot3^{n-1}\). Substitution checks it: for \(n=1\), the term is 5; for \(n=2\), it is 15; for \(n=3\), it is 45. Option A, \(5n\), gives 10 at \(n=1\), so it is not the sequence. Option C starts with 3 rather than 5 and uses the wrong base. Option D, \(15n\), is linear and does not begin with 5. Remember: in a GP the exponent is \(n-1\), because the first term has zero multiplications by the ratio.

The governing concept is the term-to-term rule of a geometric progression: each term is obtained by multiplying the preceding term by the same common ratio. Here a_1 = 18 and r = 1/3. Therefore a_2 = a_1r = 18 × 1/3 = 6, so option B is correct. This can also be checked from the general formula a_n = a_1r^(n-1): for n = 2, a_2 = 18(1/3)^1 = 6. Because the ratio is less than 1, the second term should be smaller than 18, which rules out 9 and 12 as well as any larger value. Option 3 would require an incorrect division by 6 rather than multiplication by one-third. Thus the ratio operation and the direct formula agree.

The ratios (4\div1=4) and (9\div4=\frac{9}{4}) are not equal. In exams, equal ratio is necessary for a geometric progression.

In a geometric progression, the common ratio \(r\) is found by dividing any nonzero term by the term immediately before it. Using the first two terms gives \(r=\frac{12}{4}=3\). This is confirmed by the next pairs: \(\frac{36}{12}=3\) and \(\frac{108}{36}=3\). Therefore, option B is correct. Option A would describe doubling, option C would require each term to be four times the previous one, and option D would require a factor of six. Since every displayed transition is multiplication by 3, the sequence is consistent with \(r=3\).

The \(n\)th term of a geometric progression is \(T_n=ar^{n-1}\). Hence, \(T_4=6\times2^{4-1}=6\times2^3=48\). The value 24 is the third term, since \(T_3=6\times2^2=24\). Exam tip: the exponent of the common ratio in the \(n\)th term is always \(n-1\).

The common ratio is (\frac{1}{2}), so the fifth term is (\frac{25}{2}\times\frac{1}{2}=\frac{25}{4}). In exams, simplify fractions.

The direct answer is C: 16 is the fifth term. Start with the first term and multiply by the common ratio 2 each time: first term 1, second 2, third 4, fourth 8, and fifth 16. Algebraically, \(a_n=1\cdot2^{n-1}=2^{n-1}\). Setting \(2^{n-1}=16=2^4\) gives \(n-1=4\), so \(n=5\). Option A says third term, but the third term is 4. Option B says fourth term, but the fourth term is 8. Option C is correct because the fifth term is 16. Option D says sixth term; the sixth term would be 32. Listing a short GP is safe here, because the requested number is small. The useful memory trick is to count the starting 1 as term one; do not count the number of multiplications alone, since four multiplications lead from 1 to the fifth term.

Given \(a_n=3\cdot2^{n-1}\): for \(n=1\), \(a_1=3\cdot2^0=3\); for \(n=2\), \(a_2=3\cdot2^1=6\); and for \(n=3\), \(a_3=3\cdot2^2=12\). Hence, the first three terms are \((3, 6, 12)\). The sequence \((3, 9, 27)\) has common ratio 3, whereas this sequence has common ratio 2. Exam tip: substitute \(n=1\) first and check the exponent \(n-1\) carefully.

The first term is (8) and the ratio is (3), so (a_n=8\cdot3^{n-1}). In exams, match both the first term and the ratio.

The direct answer is D: 250. In a geometric progression, each term is obtained by multiplying the previous term by the same common ratio. Divide consecutive terms: \\(10/2=5\\) and \\(50/10=5\\), so the common ratio is \\(r=5\\). The missing term follows the same multiplication: \\(50\times5=250\\). A, 100, would multiply by 2 from 50 and would break the common ratio. B, 150, would multiply by 3 and also breaks the pattern. C, 200, would multiply by 4 and is not consistent. D is correct because it preserves ratio 5; the next term would then be \\(250\times5=1250\\). The first term is not added repeatedly, as in an arithmetic progression; it is multiplied repeatedly. Memory cue: GP means “same multiplier every time.”

The ratio of consecutive terms is (16\div32=\frac{1}{2}). In exams, write the fractional ratio for a decreasing geometric progression.

The fifth term is (486), not the sixth; the asked term should be counted carefully. In exams, include the first term while counting.