Geometric Progression

For three consecutive terms p, q, and s of a geometric progression, the middle-term property is q² = ps. Applying this property to x, 3x, and 27 gives (3x)² = x·27. Thus 9x² = 27x, so 9x(x−3) = 0. The algebraic solutions are x = 0 and x = 3. Because the question specifically asks for the positive value, x = 3 must be selected. Option C is correct. Substitution verifies the result: the terms become 3, 9, and 27, and both consecutive ratios are 9/3 = 3 and 27/9 = 3. Zero is not positive, and 1, 2, and 9 fail the middle-term relation or do not produce equal consecutive ratios.

The governing concept is the nth-term formula for a geometric progression: aₙ = a₁rⁿ⁻¹. The exponent is n − 1 because the first term has undergone zero multiplications by the common ratio. With a₁ = 6 and r = 3, the fourth term is a₄ = 6×3³ = 6×27 = 162, and the sixth term is a₆ = 6×3⁵ = 6×243 = 1458. Hence a₆ − a₄ = 1458 − 162 = 1296, so option C is correct. Option D is only the value of a₆, not the requested difference. Options A and B do not result from correctly finding both indexed terms and subtracting the fourth from the sixth. Checking the sequence gives 6, 18, 54, 162, 486, 1458, which independently confirms that the required difference is 1296.

Each term is (\frac{1}{3}) of the previous term, so the progression decreases. Check both the value and direction of the ratio.

For a geometric progression, \(a_n=a_m r^{n-m}\). Hence, \(a_5=a_2r^{5-2}=24\left(\frac{1}{2}\right)^3=24\times\frac{1}{8}=3\). Note that \(a_4=6\), so 6 is a close distractor; to reach \(a_5\) from \(a_2\), multiply by the common ratio three times.

The terms are (2,10,50,250,1250,6250), so the greatest term below (6000) is (1250). The next term crosses the limit.

In a GP, dividing each term by its preceding term gives the same common ratio: \(a_{n+1}/a_n=r\). Option A describes an arithmetic progression with a constant difference. Exam tip: compare ratios of consecutive terms, not differences.

The governing concept is the general term of a geometric progression, aₙ = a₁rⁿ⁻¹, together with multiplication of the required terms. Starting from a₁ = 1 and multiplying successively by r = 4 gives a₁ = 1, a₂ = 4, a₃ = 16, and a₄ = 64. Therefore the product is 1 × 4 × 16 × 64. Since 4 × 16 = 64 and 64 × 64 = 4096, the correct answer is 4096, or option C. An efficient check is a₁⁴r^(0+1+2+3) = 1⁴ × 4⁶ = 4096. The other choices can arise from using the wrong number of powers, omitting a term, or making an arithmetic error; none equals the product of all four specified terms.

The general term of a GP is \(a_n=a_1r^{n-1}\). Thus, \(a_4=9r^3=72\), so \(r^3=8\) and the positive value of \(r\) is 2. Option 8 is the value of \(r^3\), not of \(r\). Exam tip: in \(a_n\), the exponent of \(r\) is always \(n-1\).

(\frac{a_5}{a_3}=r^2=3^2=9). In the same progression, the ratio of terms depends on the difference of positions.

The governing idea is that equal index gaps in a geometric progression correspond to equal multiplicative factors. From a₂ to a₄ there are two steps, so a₄ = a₂r². Substituting gives 80 = 20r², hence r² = 4. Since r is positive, r = 2. From a₄ to a₆ there are again two steps, so the same factor r² = 4 applies. Therefore a₆ = a₄r² = 80 × 4 = 320, making option C correct. Option A uses only one multiplication by r, option B does not follow any valid two-step factor, and option D uses an excessive factor. The positive-ratio condition removes the negative value r = −2.

In a geometric progression, the ratio of consecutive terms is constant. Here, \(\frac{48}{12}=4\). Therefore, \(\frac{12}{x}=4\), which gives \(x=3\). If \(x=4\), the two ratios would be \(3\) and \(4\), so the terms would not form a GP. Exam tip: for three consecutive GP terms \(a,b,c\), you may also use \(b^2=ac\).

The first (5) terms are (12,6,3,\frac{3}{2},\frac{3}{4}), and their sum is (\frac{93}{4}). Add fractional terms using a common denominator.

In a geometric progression, \(a_n=a_1r^{n-1}\). Therefore, \(a_3=11\times3^2=99\) and \(a_5=11\times3^4=891\). Hence, \(a_3+a_5=99+891=990\). Options such as 1000 do not follow from calculating the terms with the required powers of the common ratio. Exam tip: for the \(n\)th term, use the exponent \(n-1\).

In a geometric progression, the \(n\)th term is \(a_n=a_1r^{n-1}\). Therefore, \(a_4=8\left(\frac{3}{2}\right)^{4-1}=8\times\frac{27}{8}=27\). Option 36 may result from using an incorrect power of the common ratio or multiplying incorrectly. Exam tip: in \(a_n\), the exponent of \(r\) is always \(n-1\), not \(n\).

The governing concept is the nth-term formula for a geometric progression: aₙ = a₁rⁿ⁻¹. In this sequence, a₁ = 1 and the common ratio is r = −3, because each term is obtained by multiplying the preceding term by −3. Hence a₇ = 1 × (−3)⁶. The exponent is even, so the sign is positive, and 3⁶ = 729. Thus a₇ = 729, making option C correct. The sign pattern independently confirms this result: the first, third, fifth, and seventh terms are positive, while the even-position terms are negative. The negative options incorrectly retain a negative sign for an even power. The value 243 comes from using an incorrect exponent or incomplete power calculation, not from the seventh term.

(a_1=\frac{14}{5}) and (a_3=70), so the sum is (\frac{364}{5}). Use the ratio both backward and forward.

(x^2=16\cdot81=1296), so positive (x=36). The square of the middle term equals the product of the outer terms.

In a geometric progression, each term is obtained by multiplying the preceding term by the same constant ratio. In option B, \(18/6=3\) and \(54/18=3\), so the common ratio is 3 and the terms form a GP. In option A, the ratios are 2 and \(5/2\), so it is not a GP. Exam tip: for three terms \(a,b,c\), check whether \(b/a=c/b\).

(625=5r^3), so (r^3=125) and (r=5), hence (a_5=3125). First find the ratio and then form the next term.

The governing concept is the nth term of a geometric progression. In the sequence 7, 14, 28, 56, the first term is a = 7 and each term is multiplied by the common ratio r = 2. The nth-term formula is a_n = a × r^(n−1). Therefore, the twelfth term is a_12 = 7 × 2^11 = 7 × 2048 = 14,336. Hence option B is correct. A represents an earlier power calculation, while C and D are respectively twice and four times the correct value; they can result from using the wrong exponent or continuing the doubling one or two extra steps. The index n−1 is essential.