Geometric Progression

From the fourth term to the seventh term, the position gap is 3, so \(a_7=a_4r^3\). Thus, \(1458=54r^3\), giving \(r^3=27\) and hence \(r=3\). If the ratio were 6, then \(54\times6^3\) would not equal 1458. Exam tip: for two terms \(a_m\) and \(a_n\), use \(a_n=a_mr^{n-m}\).

In 3, 6, 12, 24, the ratios of consecutive terms are 6/3 = 12/6 = 24/12 = 2, so it is a geometric progression. In option B, the differences change. Exam tip: identify a GP by checking equal consecutive ratios.

The direct answer is Option C:
(-1215). In a geometric progression, each term is obtained by multiplying the previous term by the same common ratio. Here, divide the second term by the first:
(-15) divided by 5 equals
(-3). Check:
(-15)(-3)=45, 45(-3)=-135, so the ratio is correct. Continue carefully: first term 5; second term
(-15); third 45; fourth
(-135); fifth
405; sixth
(-1215). Equivalently, the sixth term is
(5(-3)^5)=
(-1215). Option A, 405, is the fifth term, so it stops one step too early. Option B,
(-405), has the wrong magnitude and is not produced by multiplying the fifth term by
(-3). Option C is correct because
405(-3)=-1215. Option D, 1215, has the wrong sign; the negative ratio makes the signs alternate positive, negative, positive, negative, positive, negative. A good exam habit is to find the ratio first, then write one more term instead of guessing the sign.

For three consecutive terms of a geometric progression, the square of the middle term equals the product of its neighbouring terms. Thus, \(x^2=4\times100=400\). Hence \(x=\pm20\), but the progression is stated to be positive, so \(x=20\). Taking \(40\) does not give a common ratio. Exam tip: for consecutive GP terms \(a,b,c\), use \(b^2=ac\) directly.

(a_5=a_1r^4), so (243=3r^4) and (r^4=81), hence (r=3). When positive ratio is asked, take the positive root.

The direct answer is B: 72. In a geometric progression, moving from the third term to the sixth term takes three multiplications by r, so \(a_6=a_3r^3\). Substitute the given values: \(648=24r^3\). Dividing by 24 gives \(r^3=27\), hence the real common ratio is \(r=3\). The next term is \(a_4=a_3r=24\times3=72\). Option A, 48, would correspond to a ratio of 2 from the third term. Option B is correct because it follows the calculated ratio. Option C, 96, would require a ratio of 4, and option D, 144, would require a ratio of 6; neither satisfies \(r^3=27\). A useful check is that 24, 72, 216, 648 are consecutive terms, so the sixth term is indeed reached after three factors of 3. Common mistake: use the exponent difference 6-3=3, not 6.

The direct answer is option B: (6, 18, 54, 162, …). We must check both conditions, not just one. In option B, the first term is 6, the second term is 18, so \(a_2=18\). Continuing the geometric pattern, the common ratio is 18 ÷ 6 = 3. The terms are 6, 18, 54, 162, and 486; therefore \(a_5=486\) as well. Option A is 2, 6, 18, 54, …; its second term is 6, not 18, although its fifth term would be 162. Option B satisfies both required positions. Option C is 9, 18, 36, 72, 144, …; it has \(a_2=18\), but \(a_5=144\), not 486. Option D is 18, 54, 162, 486, …; 486 is its fourth term, not its fifth, and its second term is 54. The exam lesson is to count positions carefully and verify every condition. Memory cue: write the fifth term explicitly before deciding.

Given \(a_n=2\cdot3^{n-1}\), \(a_4=2\cdot3^3=54\) and \(a_6=2\cdot3^5=486\). Hence, \(a_4+a_6=54+486=540\). Option 486 is only the value of \(a_6\), not the sum of both terms. Exam tip: substitute the term number carefully into the exponent \(n-1\).

The terms are (160,80,40,20,10,5,\frac{5}{2}), so it is the seventh term. In a decreasing progression, multiply repeatedly by the ratio.

For the middle term, ((x+4)^2=(x-2)(2x+8)), which gives (x=8). In three GP terms, the square of the middle term equals the product of the extremes.

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

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.

Direct answer: Option A, 2. A geometric progression (GP) is a sequence in which each term is obtained by multiplying the previous term by the same common ratio r. Here the first term is 2 and five terms must total 62. Test r=2: the terms are 2, 4, 8, 16, and 32. Adding step by step gives 2+4=6, +8=14, +16=30, and +32=62, exactly as required. Option A works. Option B, r=3, gives 2+6+18+54+162=242, not 62. Option C, r=4, gives 2+8+32+128+512=682, not 62. Option D, r=5, gives 2+10+50+250+1250=1562, not 62. Thus only A is correct. Memory cue: for a GP, keep multiplying by r and then add the required terms.

(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 \(n-1\).

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 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 \(n-1\).

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.