Geometric Progression

For three consecutive terms \(a,b,c\) of a geometric progression, \(b^2=ac\). Thus, \(x^2=5\times45=225\), giving \(x=\pm15\). Since the progression is positive, \(x=15\). For example, with \(x=25\), \(25/5\) and \(45/25\) are not equal, so the terms do not form a GP. Exam tip: for three consecutive GP terms, the square of the middle term equals the product of the outer terms.

The sum of the first four terms is (15+30+60+120=225), and the average is (\frac{225}{4}), so none of the given options is correct. In average questions, first find the sum.

The third term of a GP is \(a_3=a_1r^{3-1}\). Therefore, \(a_3=6\times4^2=6\times16=96\). The value 72 would result from applying the ratio only once, which gives the second term. Exam tip: use \(a_n=a_1r^{n-1}\) to find the \(n\)th term of a GP.

The common ratio is (\frac{25}{125}=\frac{1}{5}). In a decreasing geometric progression the ratio can be less than (1).

Given \(a_n=5\cdot3^{n-1}\), substitute \(n=5\): \(a_5=5\cdot3^{5-1}=5\cdot3^4=5\cdot81=405\). Hence, 405 is correct. The value 135 may result from incorrectly using \(3^3\). Exam tip: substitute the term number first and simplify the exponent \(n-1\) carefully.

Direct answer: 256 is the sixth term, so option C is correct. Each term is twice the preceding term, so the common ratio is 2. Count the terms carefully: first 8, second 16, third 32, fourth 64, fifth 128, and sixth 256. The general-term equation gives the same result: aₙ = 8 × 2ⁿ⁻¹. Setting aₙ = 256 gives 8 × 2ⁿ⁻¹ = 256, so 2ⁿ⁻¹ = 32 = 2⁵. Therefore n − 1 = 5 and n = 6. Option A is 64, the fourth term. Option B is 128, the fifth term. Option C is 256, the sixth term. Option D would be 512, the seventh term. A common counting mistake is to call 8 the zeroth term; in this question the first displayed term is term 1.

The first term is (10), and multiplying by (2) each time gives (10,20,40,80). In a geometric progression each next term is formed by multiplication.

In a geometric progression, the common ratio is found by dividing any term by the preceding term. Here, \(r=\frac{-14}{7}=-2\). This is confirmed by \(\frac{28}{-14}=-2\) and \(\frac{-56}{28}=-2\), so the common ratio is \(-2\). Option 2 is a close distractor, but it does not account for the alternating signs of the terms. Exam tip: Always include the sign when calculating a common ratio.

The common ratio is (5), so the next term is (150\cdot5=750). Multiply the previous term by the ratio to get the next 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, \(x^2=9\times81=729\). Hence \(x=\sqrt{729}=27\), since the terms are stated to be positive. A value such as \(18\) does not give a common ratio. Exam tip: for consecutive GP terms \(a,b,c\), use \(b^2=ac\).

The first four terms are 8, 24, 72, and 216. Therefore, \(8+24+72+216=320\). Hence, 320 is the correct answer. Leaving out 216 would give 104, which is the sum of only the first three terms. Exam tip: When only a few terms are given, add every listed term directly to verify the answer.

In a geometric progression, the second term is \(a_2=a_1r\). Thus, \(28=a_1\times4\), so \(a_1=28\div4=7\). Multiplying 28 by 4 gives 112, which would be the next term \(a_3\), not the first term. Exam tip: To find a previous term in a GP, divide by the common ratio.

The general term of a GP is \(a_n=a_1r^{n-1}\). Therefore, \(a_4=3r^3=81\), so \(r^3=27\). Since the common ratio is positive, \(r=3\). If \(r=9\), the fourth term would be \(3\times9^3\), not 81. Exam tip: in \(a_n\), the exponent of \(r\) is always \(n-1\).

The governing concept is a geometric progression, in which every term is obtained by multiplying the preceding term by the same constant ratio. Divide the second term by the first: \(r=(5/3)\div(5/6)=2\). The pattern is confirmed because \((10/3)\div(5/3)=2\) and \((20/3)\div(10/3)=2\). Therefore the next term is \(a_5=(20/3)\times2=40/3\). The general formula gives the same result: \(a_n=ar^{n-1}\), so \(a_5=(5/6)2^4=40/3\). Hence option B is correct. Option A does not continue the constant-ratio pattern, while C and D result from multiplying by an incorrect power or making an arithmetic error with the fractions.

The common ratio is (\frac{1}{3}), so the terms are (243,81,27,9,3,1). In a decreasing progression multiply by the ratio repeatedly.

In option A, consecutive ratios are \(-4/2=-2\), \(8/(-4)=-2\), and \((-16)/8=-2\). Hence it is a GP. In B, the ratios change. Exam tip: check successive ratios, not successive differences.

The direct answer is option B: (81, 27, 9, 3, …). The common ratio is found by dividing a term by the immediately preceding term. In option B, 27 ÷ 81 = \(\frac{1}{3}\), 9 ÷ 27 = \(\frac{1}{3}\), and 3 ÷ 9 = \(\frac{1}{3}\). Thus the ratio is consistently \(\frac{1}{3}\). Option A has ratio 9 ÷ 3 = 3, so it increases by 3 rather than decreases by a third. Option B has the required ratio and is correct. Option C has ratio 3 ÷ 1 = 3, not \(\frac{1}{3}\). Option D is not a geometric progression because 9 ÷ 27 = \(\frac{1}{3}\), but 6 ÷ 9 = \(\frac{2}{3}\); the ratio changes. Always check at least two consecutive divisions, because one matching division alone is not enough. Memory cue: a ratio of \(\frac{1}{3}\) means each next term is one-third of the previous term.

A geometric progression has a fixed ratio between consecutive terms. Here \(r=25/5=5\), and the sequence is formed by multiplying by 5 at every step. Thus \(a_1=5\), \(a_2=25\), \(a_3=125\), and \(a_4=625\). One more multiplication gives \(a_5=625\times5=3125\), so the required number is the fifth term. Using the formula \(a_n=ar^{n-1}\), we obtain \(3125=5\times5^{n-1}=5^5\), hence \(n-1=4\) and \(n=5\). Therefore option B is correct. Option A stops at 625, whereas options C and D count one or two extra multiplications and therefore represent larger terms.

Using \(a_n=27\left(\frac{1}{3}\right)^{n-1}\), substitute \(n=4\): \(a_4=27\left(\frac{1}{3}\right)^{4-1}=27\left(\frac{1}{3}\right)^3=27\times\frac{1}{27}=1\). Therefore, the correct answer is 1. Option 3 would result from incorrectly using exponent 2, but here \(n-1=3\). Exam tip: write \(n-1\) separately before evaluating the power.

The direct answer is option C: x = 64. In a geometric progression, every term is obtained by multiplying by the same common ratio. From the first two terms, the ratio is 16 ÷ 4 = 4. Therefore the third term is x = 16 × 4 = 64. Checking the last step, 64 × 4 = 256, so the complete sequence is consistent. Option A, 32, would make the ratio 32 ÷ 16 = 2 and would give 128 next, not 256. Option B, 48, gives a ratio of 3 and would give 144 next. Option C, 64, gives ratio 4 on both steps and is correct. Option D, 96, gives ratio 6 and would give 576 next. The key is not to guess from the size of the answer; calculate the fixed multiplier first. Memory cue: find r from the first two terms, then multiply the second term by r.

The first four terms are 14, 28, 56, and 112. Therefore, their sum is \(14+28+56+112=210\). Hence, 210 is the correct answer. Note that 224 is not the sum; it would be the next term of the progression. Exam tip: For a GP with only a few terms, direct addition is often the quickest method.

In \(2, 6, 18, 54\), the consecutive ratios are \(6/2=3\), \(18/6=3\), and \(54/18=3\), so it is a GP. In option C, the ratios are not constant. Exam tip: verify the ratio of every pair of consecutive terms.

The first term is (12) and the ratio is (3), so (a_n=12\cdot3^{n-1}). Use both first term and ratio in the general term.

The governing concept is the sum of the first n terms of a geometric progression. The first term is a = 3, the common ratio is r = 2, and n = 6. The terms are 3, 6, 12, 24, 48, and 96, whose direct sum is 3 + 6 + 12 + 24 + 48 + 96 = 189. The geometric-series formula gives the same result: for r ≠ 1, Sₙ = a(rⁿ − 1)/(r − 1). Hence S₆ = 3(2⁶ − 1)/(2 − 1) = 3(64 − 1) = 3 × 63 = 189. Therefore option A is correct. The other values can arise from an addition mistake, an incorrect power, or counting a different number of terms.

The governing rule for the nth term of a geometric progression is \(a_n=a_1r^{n-1}\). For the third term, the ratio is used twice because the sequence moves from the first term to the second and then from the second to the third. Hence \(a_3=30(1/3)^{3-1}=30(1/3)^2=30/9=10/3\). Directly, the first next term is \(30\times1/3=10\), and the following term is \(10\times1/3=10/3\), confirming the result. Therefore option A is correct. Option C is only the second term, option D is an unsimplified expression equal to the same value but is not the intended simplified choice, and option B does not follow the repeated ratio.