Geometric Progression

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.

The common ratio is \(-\frac{1}{2}\), so the next term is \((-6)\left(-\frac{1}{2}\right)=3\). With a negative ratio the signs alternate.

Given \(a_n=7\cdot2^{n-1}\), \(a_4=7\cdot2^3=56\) and \(a_5=7\cdot2^4=112\). Therefore, \(a_4+a_5=56+112=168\). Option 112 is only the value of \(a_5\), not the required sum. Exam tip: substitute the value of \(n\) carefully in the formula for each term.

The defining feature is a constant ratio: each term is twice the preceding term, so \(a=3\) and \(r=2\). For a geometric progression, the nth term is \(a_n=ar^{n-1}\). Hence \(a_{12}=3\times2^{11}=3\times2048=6144\). The exponent is 11 rather than 12 because the first term already contains the initial factor 3; reaching the twelfth position requires eleven successive multiplications by 2. Thus option B is correct. Option A corresponds to a smaller power, while options C and D result from using an incorrect initial factor or exponent. Substitution into the formula and repeated doubling both verify the answer.