Geometric Progression

Dividing \(a_n^2=a_{n-1}a_{n+1}\) by \(a_{n-1}a_n\) gives \(\frac{a_n}{a_{n-1}}=\frac{a_{n+1}}{a_n}\). Thus consecutive terms have the same ratio, so the sequence is a GP. Option A describes an AP with a constant difference. Exam tip: test the ratio, not the difference, for a GP.

The governing concept is the nth-term formula for a geometric progression: a_n = a_1r^(n−1). Here a_1 = 8 and the common ratio is r = −16/8 = −2. For the seventh term, a_7 = 8(−2)^(7−1) = 8(−2)^6. The exponent is even, so (−2)^6 = 64, giving a_7 = 8 × 64 = 512. Hence option C is correct. The signs alternate because the ratio is negative: odd-position terms are positive and even-position terms are negative, so the seventh term must be positive. Options A and B have an incorrect magnitude, while D has the wrong sign despite having the correct magnitude pattern.

For three consecutive terms of a geometric progression, the square of the middle term equals the product of the first and third terms. Thus, \(x^2=6\times150=900\). Hence \(x=\pm30\), but the progression is stated to be positive, so \(x=30\). A value such as \(36\) does not give equal common ratios. Exam tip: for consecutive GP terms \(a,b,c\), use \(b^2=ac\) directly.

(a_5=a_1r^4), so (1792=7r^4) and (r^4=256), hence (r=4). When positive ratio is asked, take the positive root.

For consecutive GP terms, the common-ratio condition gives \(q/p=r/q\). Cross-multiplying gives \(q^2=pr\). The relation \(p+r=2q\) belongs to an AP. Exam tip: square the middle term and compare it with the product of the outer terms.

In (5,15,45,135,405,1215,\ldots), (a_3=45) and (a_6=1215). Check each option up to the required positions.

Given \(a_n=4\cdot3^{n-1}\), \(a_4=4\cdot3^{3}=108\) and \(a_6=4\cdot3^{5}=972\). Therefore, \(a_4+a_6=108+972=1080\). The option 972 is only the value of \(a_6\), not the required sum. Exam tip: while finding \(a_n\), remember that the exponent is \(n-1\).

The terms are (384,192,96,48,24,12,6), so (6) is the seventh term. In a decreasing progression, multiply repeatedly by the ratio.

(a_7=a_3r^4), so (576=36r^4) and (r=2), hence (a_1=\frac{36}{2^2}=9). Divide by the ratio to move backward.

For the middle term, ((3x)^2=(x+2)(6x+4)), which gives (x=4). In three GP terms, the square of the middle term equals the product of the extremes.

The terms are (9,18,36,72,144,288,576,1152), so the greatest term below (600) is (576). In boundary questions, check the next term too.

The governing relation is a_n = a_1r^(n−1). With r = 3, the second term is a_2 = a_1 × 3 = 3a_1. The fourth term is a_4 = a_1 × 3^3 = 27a_1 because three ratio-steps separate the first and fourth terms. Substituting into the given condition gives 3a_1 + 27a_1 = 90, so 30a_1 = 90 and a_1 = 3. Therefore option B is correct. A common mistake is to use 3^2 for the fourth term or to treat the sequence as arithmetic. Checking the result, the first terms are 3, 9, 27, 81 and 9 + 81 = 90, confirming the answer.

In (32,16,8,4,2,\ldots), (r=\frac{1}{2}) and the fifth term is (2). Check both first and fifth terms in the options.

For (r=3), the terms are (4,12,36,108,324) and the sum is (484). For small options, forming the terms is quick.

The first term is \(a=5\), and the common ratio is \(r=\frac{-20}{5}=-4\). Thus, \(a_3=5(-4)^2=80\) and \(a_5=5(-4)^4=1280\). Therefore, \(a_5+a_3=1280+80=1360\). An answer such as \(1120\) can result from using the wrong power or term position with the negative common ratio. Exam tip: use \(a_n=ar^{n-1}\) and check the exponent before calculating.

Given \(a_n=96\left(\frac{1}{2}\right)^{n-1}\), \(a_3=96\left(\frac{1}{2}\right)^2=24\) and \(a_6=96\left(\frac{1}{2}\right)^5=3\). Therefore, \(a_3+a_6=24+3=27\). The option 30 may result from using an incorrect exponent instead of \(n-1\). Exam tip: write the exponent \(n-1\) first before evaluating each term.

For a geometric progression, moving from a_2 to a_6 involves four ratio-steps, so a_6 = a_2r^4. Substituting the given values gives 8192 = 32r^4, hence r^4 = 256. The positive-r condition gives r = 4 rather than −4. Since a_2 = a_1r, we have 32 = 4a_1, so a_1 = 8. Therefore option C is correct. The positivity condition is essential: without it, a negative ratio could also satisfy the fourth-power equation, although the requested first term would then differ in sign. Options A, B and D arise from using the wrong exponent or dividing by the wrong ratio step.

The governing property is that consecutive terms of a geometric progression have a constant ratio. For three consecutive terms, the square of the middle term equals the product of its neighbors, so x^2 = 5 × 80 = 400. Because all terms are positive, x = 20 rather than −20. The common ratio is therefore r = 20/5 = 4. Multiplying the third term by the same ratio gives y = 80 × 4 = 320. Thus option C is correct. The sequence 5, 20, 80, 320 confirms that the ratio is 4 in every step. The other options fail to preserve a constant ratio between the third and fourth terms or do not agree with the positive middle term.

A geometric progression has the same ratio between every pair of consecutive terms, and a_4 means the fourth listed term. In option A, the sequence is 12, 24, 48, 96, so each term is twice the preceding term and r = 2. Its fourth term is exactly 96, satisfying both conditions. Option B has ratio 2 but its fourth term is 192. Option C also has ratio 2, but its fourth term is 48. Option D has ratio 2 as well, yet its fourth term is 768. Therefore only option A is correct. The question requires checking both the common ratio and the position of the specified term, not just one of them.

The nth term of a geometric progression is \(a_n=a_1r^{n-1}\). Therefore, \(a_4=\frac{7}{2}\times4^{4-1}=\frac{7}{2}\times64=224\). The value \(448\) usually results from an error in evaluating the power of \(4\) or in multiplication. Exam tip: for the fourth term, the exponent of the common ratio is always \(4-1=3\).