Geometric Progression

The terms are (3,6,12,24,48,96,192), so the greatest term below (100) is (96). In boundary questions, check nearby terms.

Use the geometric-progression relation between separated terms. From the second term to the sixth term there are four ratio steps, so \(a_6=a_2r^4\). Substitution gives \(160=10r^4\), hence \(r^4=16\). Since the ratio is positive, \(r=2\), not the negative fourth-root possibility. Now the second term satisfies \(a_2=a_1r\), so \(10=2a_1\), which gives \(a_1=5\). Therefore option D is correct. Verification is immediate: the sequence begins \(5,10,20,40,80,160\). The other choices do not produce both the stated second term and the stated sixth term under the same positive ratio.

In (12,6,3,\frac{3}{2},\ldots), (r=\frac{1}{2}) and the fourth term is (\frac{3}{2}). Check both first and fourth terms in the options.

For (r=2), the terms are (2,4,8,16,32) and the sum is (62). For small options, forming the terms is quick.

(a_4=-243) and (a_6=-2187), so the sum is (-2430). The correct option should include (-2430).

Using \(a_n=48\left(\frac{1}{2}\right)^{n-1}\), \(a_3=48\left(\frac{1}{2}\right)^2=12\) and \(a_5=48\left(\frac{1}{2}\right)^4=3\). Hence, \(a_3+a_5=12+3=15\). The option 18 can result from using an incorrect exponent or term number. Exam tip: after substituting \(n\), always use the exponent \(n-1\).

The governing concept is the relationship between terms of a geometric progression: a_m = a_k r^(m−k). Moving from the second term to the fifth term requires three multiplications by r, so a₅ = a₂r³. Substituting the given values gives 405 = 15r³, hence r³ = 27. Since r is specified as positive, r = 3. Now use a₂ = a₁r: 15 = 3a₁, so a₁ = 5. Therefore option C is correct. The value 3 is the common ratio, not the first term. Values 4 and 6 do not produce both a₂ = 15 and a₅ = 405 with one positive common ratio, so they are not consistent with the progression.

In a geometric progression, the square of a middle term equals the product of the two terms equally far from it. For the four terms \\(2,x,18,y\\), the first three terms show that \\(x\\) is the positive geometric mean of 2 and 18. Positivity is important because it selects the positive square root.

Thus \\(x^2=2\cdot18=36\\), so \\(x=6\\), not \\(-6\\). The common ratio is then \\(r=6\div2=3\\). Multiplying the third term by this ratio gives \\(y=18\cdot3=54\\). Therefore option C is correct. The value 36 is only an intermediate product, not the required fourth term.

The governing concept is the definition of a geometric progression: every term must be obtained by multiplying the preceding term by the same common ratio. We must check two independent conditions, not just one. In option A, the sequence is 2, 10, 50, 250, so its third term is 50. Also, 10/2 = 5, 50/10 = 5, and 250/50 = 5; therefore its common ratio is 5. Option B has ratio 5 but its third term is 125. Option C also has ratio 5, but its third term is 250. Option D has ratio 5, but its third term is 25. Hence only option A satisfies both requirements, so A is unambiguously correct.

The nth term of a geometric progression is \(a_n=a_1r^{n-1}\). Therefore, \(a_5=\frac{3}{2}\times4^{5-1}=\frac{3}{2}\times4^4=\frac{3}{2}\times256=384\). Hence, 384 is correct. Option 192 results from an incorrect multiplication involving the first term \(\frac{3}{2}\). Exam tip: in \(a_n\), the exponent of the common ratio is always \(n-1\), not \(n\).

The terms are (1,3,9,27,81,243,729,2187), so the greatest term below (1000) is (729). In boundary questions, check the next term too.

In a GP, \(b=ar\) and \(c=br\), so \(b^2=(ar)^2=a(br)=ac\). Conversely, \(b^2=ac\) gives \(b/a=c/b\). Exam tip: the square of the middle term equals the product of the extreme terms.

The first term of this GP is 6 and the common ratio is 3. Thus, \(a_3=6\times 3^2=54\) and \(a_5=6\times 3^4=486\). Therefore, \(a_5-a_3=486-54=432\). Option 486 is only the fifth term, not the required difference. Exam tip: use \(a_n=ar^{n-1}\) to find the required terms of a GP.

(a_4=a_2r^2), so (r^2=16), and for the positive sequence (a_6=128\cdot16=2048). The same (r^2) applies over the same gap.

Given \(a_n=81\left(\frac{1}{3}\right)^{n-1}\), \(a_2=81\left(\frac{1}{3}\right)^1=27\) and \(a_5=81\left(\frac{1}{3}\right)^4=1\). Hence, \(a_2+a_5=27+1=28\). Option 30 may result from incorrectly taking \(a_5\) as 3. Exam tip: while substituting a term number, carefully use the exponent \(n-1\).

The governing concept is the finite sum of a geometric progression, while the negative ratio requires careful sign handling. Starting with a₁ = 4 and multiplying successively by −2 gives the first five terms: 4, −8, 16, −32, and 64. Their signed sum is 4 − 8 + 16 − 32 + 64 = 44. The formula confirms this result: S₅ = a₁(1−r⁵)/(1−r) = 4[1−(−2)⁵]/[1−(−2)] = 4(33)/3 = 44. Thus option D is correct. Adding absolute values would incorrectly give a different result because the signs alternate. Options A, B, and C can arise from ignoring a negative sign, omitting a term, or making an arithmetic error.

In a geometric progression, \(a_7=a_3r^{7-3}=a_3r^4\). Thus, \(3645=45r^4\), giving \(r^4=81\). Since \(r\) is positive, \(r=3\). Although \(r=-3\) also satisfies \(r^4=81\), it is excluded by the given positive condition. Exam tip: the exponent of the common ratio equals the difference between the term numbers.

In (32,-16,8,-4,\ldots), each term is multiplied by (-\frac{1}{2}). Divide the next term by the previous term to find the ratio.

In a GP, \(a_4=a_2r^2\). Thus, \(54=6r^2\), so \(r^2=9\) and \(r\) can be \(3\) or \(-3\). For \(r=3\), \(a_1=2\) and \(a_5=162\), giving \(164\). However, for \(r=-3\), \(a_1=-2\) and \(a_5=-162\), giving \(-164\). Therefore, the information given does not determine a unique value. Exam tip: when finding \(r\) from \(r^2\), check both positive and negative roots.

Given \(a_n=5\cdot2^{n-1}\) and \(a_n=320\), we get \(5\cdot2^{n-1}=320\). Thus, \(2^{n-1}=64=2^6\). Equating exponents gives \(n-1=6\), so \(n=7\). If \(n=6\), the term would be \(5\cdot2^5=160\), not 320. Exam tip: first divide by the constant factor to isolate the exponential part.