Geometric Progression

The common ratio is (\frac{40}{8}=5), so (x=40\cdot5=200). Multiply the second term by the ratio to get the third term.

This is a geometric progression because every term is obtained by multiplying the preceding term by the same number. Here, the common ratio is 4: 24 divided by 6 is 4, 96 divided by 24 is 4, and 384 divided by 96 is 4. Therefore, the terms grow as 6, 24, 96, 384, and so on.

Starting from the first term, the fifth term is obtained by multiplying by 4 four times. Thus the terms in order are first term 6, second term 24, third term 96, fourth term 384, and fifth term 1536. Hence 1536 is the fifth term, so option B is correct. The answer is not the fourth term because 384 is the fourth term.

The governing property is that terms of a geometric progression separated by several positions differ by the corresponding power of the common ratio. From the second term to the fifth term there are three steps, so \(a_5=a_2r^{5-2}=16r^3\). Substitution gives \(1024=16r^3\). Dividing by 16 yields \(r^3=64\), and the positive cube root is \(r=4\). A check gives \(16\times4^3=16\times64=1024\). Thus option C is correct. If \(r=2\), the result would be 128; if \(r=3\), it would be 432; and \(r=8\) would produce a much larger value. The positivity condition also makes the intended positive root explicit.

The defining concept is a geometric progression with a possibly negative common ratio. Dividing consecutive terms gives \(r=-15/5=-3\); this is confirmed because \(45/(-15)=-3\) and \((-135)/45=-3\). To obtain the fifth term, multiply the fourth term by the ratio: \(a_5=(-135)(-3)=405\). Using the formula gives the same result: \(a_5=5(-3)^4=5\times81=405\). The two negative factors produce a positive fifth term, which is why option A is correct. Option B ignores the sign change at the last multiplication, while C and D arise from using an incorrect magnitude or exponent.

The first term is (7), and multiplying by (4) each time gives (7,28,112,448). Match both (a) and (r).

The first three terms are 6, 18, and 54. Their product is \(6\times18\times54=108\times54=5832\). Hence, 5832 is correct. A value such as 3888 can result from an incorrect multiplication. Exam tip: multiply in steps—first find \(6\times18\), then multiply the result by 54.

The common ratio is (\frac{1}{3}), so the next term is (10\cdot\frac{1}{3}=\frac{10}{3}). Multiply by the ratio.

The formula gives the term corresponding to any positive integer value of \(n\). To identify the sequence, substitute the first few values of \(n\), beginning with \(n=1\). The exponent \(n-1\) is zero for the first term, so the first term is 8; every increase of one in \(n\) doubles the value.

For \(n=1,2,3,4\), the formula gives \(8\cdot2^0=8\), \(8\cdot2^1=16\), \(8\cdot2^2=32\), and \(8\cdot2^3=64\). Thus the sequence is \((8,16,32,64,\ldots)\), which matches option C. Option B starts with 2 and follows a different multiplier, while option D incorrectly begins with 16.

In a geometric progression, the ratio of consecutive terms is constant. Here, the common ratio is \(\frac{100}{20}=5\). Hence \(\frac{20}{x}=5\), so \(x=\frac{20}{5}=4\). Option 5 is the common ratio, not the first term. Exam tip: for three consecutive GP terms \(a,b,c\), you may also use \(b^2=ac\).

In this GP, the first term is \(a=6\) and the common ratio is \(r=18/6=3\). The \(n\)th term is \(a_n=ar^{n-1}\). Therefore, \(a_6=6\times3^{5}=6\times243=1458\). Getting 729 usually results from using an incorrect power or term number. Exam tip: for the \(n\)th term, the exponent of the common ratio is always \(n-1\).

In (2,12,72,432,\ldots), each term is multiplied by (6). For the ratio divide the next term by the previous term.

In a geometric progression, the nth term is \(a_n=a_1r^{n-1}\). Therefore, \(a_4=\frac{7}{4}\times2^{4-1}=\frac{7}{4}\times8=14\). \(\frac{7}{2}\) is only the second term, so it is not correct. Exam tip: in \(a_n\), the exponent of the common ratio is always \(n-1\).

A geometric progression has a constant ratio between consecutive terms. Here 8 ÷ 4 = 2, 16 ÷ 8 = 2, and 32 ÷ 16 = 2, so the first term is a = 4 and the common ratio is r = 2. The first six terms are 4, 8, 16, 32, 64, and 128. Their sum is 4 + 8 + 16 + 32 + 64 + 128 = 252. Using the formula Sₙ = a(rⁿ − 1)/(r − 1), we obtain S₆ = 4(2⁶ − 1)/(2 − 1) = 4(64 − 1) = 4 × 63 = 252. Therefore option C is correct. The number 256 is a power-related distractor, not the sum of these six terms; the other choices do not satisfy the direct addition.

A geometric progression is generated recursively by multiplying each term by the common ratio. Begin with \(a_1=5\). The second term is \(a_2=5(-2)=-10\); the third is \(a_3=(-10)(-2)=20\); and the fourth is \(a_4=20(-2)=-40\). Thus the first four terms are \(5,-10,20,-40\), so option B is correct. Because the ratio is negative, the signs alternate at every step, while the absolute values double. Option A incorrectly treats the ratio as positive, option C lists the ratio-related values instead of repeatedly multiplying the starting term, and option D begins with the wrong first term.

For three consecutive terms of a geometric progression, the square of the middle term equals the product of the first and third terms. Thus, \(x^2=16\times81=1296\). Hence \(x=\pm36\), but the progression is positive, so \(x=36\). Indeed, 16, 36, 81 has common ratio \(\frac{9}{4}\). Option 32 cannot be the middle term because \(32^2\neq16\times81\). Exam tip: for consecutive GP terms \(a,b,c\), use \(b^2=ac\) directly.

The first term is (13) and the ratio is (3), so (a_n=13\cdot3^{n-1}). Use the first term and ratio in the general term.

In a geometric progression, moving from the second term to the fifth term involves three multiplications by the common ratio. Consequently, \(a_5=a_2r^3\). Substituting the data gives \(320=40r^3\). After division by 40, we get \(r^3=8\). Since the problem specifies a positive ratio, the correct real value is \(r=2\). Direct verification is \(40\to80\to160\to320\), which uses three multiplications by 2 and reaches the given fifth term exactly. Therefore option A is correct. The alternatives 3, 4, and 8 have cubes 27, 64, and 512 respectively, so none satisfies the required equation \(r^3=8\).

Direct answer: a₄ = 625, so option C is correct. The consecutive quotients are 25/5 = 5 and 125/25 = 5, so the common ratio is 5. The first term is also 5. Apply the nth-term formula aₙ = arⁿ⁻¹: a₄ = 5 × 5³ = 5 × 125 = 625. Building the sequence gives the same result: 5, 25, 125, and then 125 × 5 = 625. Option A and option B do not follow multiplication by the fixed ratio 5. Option C is the fourth term. Option D is 3125, which is the fifth term because it is 625 × 5; it is one step too far. The useful memory cue is that a₄ requires three multiplications after the first term, so the exponent is 4 − 1 = 3, not 4.

(a_3=a_1r^2), so (150=6r^2) and (r^2=25), hence positive (r=5). When positive ratio is asked take the positive root.

Each term is half of the previous term, so (r=\frac{1}{2}). To check the ratio divide the next term by the previous term.

This is a geometric progression with first term \(a=3\) and common ratio \(r=4\). Therefore, \(a_n=3\times4^{n-1}\). From \(3\times4^{n-1}=768\), we get \(4^{n-1}=256=4^4\). Hence, \(n-1=4\) and \(n=5\). Option 4 is incorrect because the fourth term is \(192\). Exam tip: Use \(a_n=ar^{n-1}\) to find the position of a term in a GP.

In the given geometric progression, the second term is \(a_2=18\) and the fourth term is \(a_4=162\). Therefore, \(a_2+a_4=18+162=180\). Hence, option B is correct. A value such as 184 can result from identifying a term incorrectly or making an addition error. Exam tip: always count the first term as \(a_1\).

The governing concept is the finite sum of a geometric progression. Starting with 32 and multiplying repeatedly by 1/2 gives the first four terms 32, 16, 8, and 4. Their sum is 32 + 16 + 8 + 4 = 60, so option D is correct. The standard formula provides an independent check: Sₙ = a(1 − rⁿ)/(1 − r). Thus S₄ = 32[1 − (1/2)⁴]/[1 − 1/2] = 32(15/16)/(1/2) = 60. The other choices do not equal the total of all four terms; they result from omitting a term, using an incorrect power, or making an addition error. Both methods agree.

In a geometric progression, moving from the third term to the sixth term requires three multiplications by the common ratio. Therefore a_6=a_3r^{6-3}=20(3^3). Since 3^3=27, the product is 20times27=540. Thus the sixth term is 540.

Option C is correct. The exponent is not 6, because the known term is already the third term; only the three steps from positions 3 to 6 matter. Using just one multiplication would give 60, and using two would give 180, which is option A but corresponds only to the fifth term. The position formula prevents this off-by-one error and confirms the exact value 540.

For three consecutive terms of a geometric progression, the square of the middle term equals the product of the first and third terms: \(x^2=12\times75=900\). Thus \(x=\pm30\), but the progression is positive, so \(x=30\). If \(45\) is used, the two consecutive ratios are not equal. Exam tip: for consecutive GP terms \(a,b,c\), use \(b^2=ac\) directly.