Geometric Progression

The terms are (3,15,75,375,1875,9375,46875), so the greatest term below (10000) is (9375). In boundary questions, check the next term too.

In a geometric progression, \(a_n=a_1r^{n-1}\). Thus, \(a_4=9\times3^3=243\) and \(a_5=9\times3^4=729\). Therefore, \(a_4+a_5=243+729=972\). A value such as \(1215\) can result from using an incorrect power or term value. Exam tip: for the \(n\)th term, use the exponent \(n-1\).

The first term is 12 and the common ratio is 3. Thus, \(a_4=12\times 3^3=324\) and \(a_6=12\times 3^5=2916\). Therefore, \(a_6-a_4=2916-324=2592\). A value such as 2484 can result from an error in subtraction. Exam tip: while using \(a_n=ar^{n-1}\) for a GP, remember that the exponent is \(n-1\).

The governing relation is a_(m+k) = a_mr^k. From the third term to the fifth term there are two steps, so a_5 = a_3r^2. Substitution gives 288 = 18r^2, hence r^2 = 16. Since r is specified as positive, r = 4. From the fifth term to the seventh term there are again two steps, so a_7 = a_5r^2 = 288 × 16 = 4608. Therefore option C is correct. It is not necessary to calculate every intermediate term; the equal two-step gaps allow the same multiplier r^2 to be used twice. The other options result from using r instead of r^2 or from an arithmetic error.

Given \(a_n=125\left(\frac{1}{5}\right)^{n-1}\), \(a_2=125\left(\frac{1}{5}\right)^1=25\) and \(a_4=125\left(\frac{1}{5}\right)^3=1\). Hence, \(a_2+a_4=25+1=26\). The option 30 is incorrect because \(a_4\) is 1, not 5. Exam tip: for a particular term, substitute the value of \(n\) first and then evaluate the exponent carefully.

The governing concept is the finite sum of a geometric progression, Sₙ = a₁(1 − rⁿ)/(1 − r), which is valid because r ≠ 1. Substituting a₁ = 10, r = −2, and n = 5 gives S₅ = 10[1 − (−2)⁵]/[1 − (−2)] = 10(1 + 32)/3 = 330/3 = 110. Direct expansion confirms the result: the five terms are 10, −20, 40, −80, and 160, so their sum is 10 − 20 + 40 − 80 + 160 = 110. Hence option C is correct. The negative ratio makes signs alternate, but the odd power (−2)⁵ is negative, which changes the numerator to 33. Treating the ratio as positive or ignoring that sign produces distractor values.

In a geometric progression, a gap of four indices gives \,a_8=a_4r^{8-4}=a_4r^4\,. Thus \,4374=54r^4\, and \,r^4=4374/54=81\,. Hence \,r=3\, because r is stated to be positive. Although 9 may seem tempting, it is not r; it is related to the value of \,r^4\,. Exam tip: use the difference between the term indices to determine the exponent of r.

In (64,-16,4,-1,\ldots), each term is multiplied by (-\frac{1}{4}). Divide the next term by the previous term to find the ratio.

In a GP, \(a_4=a_2r^2\). Thus, \(250=10r^2\), so \(r^2=25\). Since \(r\) is positive, \(r=5\). Now \(a_1=a_2/r=10/5=2\), and \(a_5=a_4r=250\times5=1250\). Therefore, \(a_1+a_5=2+1250=1252\). Although \(r=-5\) also gives \(r^2=25\), the question specifically requires positive \(r\). Exam tip: use the difference between term indices to determine the required power of \(r\).

Given \(8\cdot3^{n-1}=1944\). Dividing both sides by 8 gives \(3^{n-1}=243\). Since \(243=3^5\), we get \(n-1=5\), hence \(n=6\). If 5 were chosen, the exponent would be \(n-1=4\), so the term would not be 1944. Exam tip: first remove the coefficient, then compare exponents with the same base.

For three consecutive terms of a geometric progression, the square of the middle term equals the product of the first and third terms. Therefore (5x)^2 = x × 125. Expanding gives 25x^2 = 125x, so 25x(x − 5) = 0. The algebraic solutions are x = 0 and x = 5, but the question asks for positive x, so only x = 5 is acceptable. Checking directly, the terms become 5, 25 and 125, and the ratio is 5 from one term to the next. Thus option C is correct. The zero solution must be rejected because zero is neither positive nor suitable for the intended nonzero geometric progression.

This question applies the nth-term rule of a geometric progression, aₙ = a₁rⁿ⁻¹. Since a₁ = 10 and r = 3, the fourth term is a₄ = 10×3³ = 10×27 = 270. The sixth term is a₆ = 10×3⁵ = 10×243 = 2430. Therefore, the required difference is a₆ − a₄ = 2430 − 270 = 2160, so option A is correct. Careful indexing is essential: the fourth term uses the third power of 3, and the sixth term uses the fifth power. Writing only 2430 gives a₆ rather than the requested difference. The other choices can arise from using an incorrect exponent, multiplying the terms instead of subtracting them, or making an arithmetic error. Listing the sequence, 10, 30, 90, 270, 810, 2430, also verifies both indexed values and the final subtraction.

Each term is (\frac{1}{2}) of the previous term, so the progression decreases. Check both the value and direction of the ratio.

\(a_6=a_3r^3=45\left(\frac{1}{3}\right)^3=\frac{5}{3}\). Moving three positions forward uses \(r^3\).

The terms are (11,22,44,88,176,352,704,1408), so the greatest term below (800) is (704). The next term crosses the limit.

For a geometric progression, the relation between two terms is determined by the common ratio. From the second term to the sixth term there are four equal steps, so \\(a_6=a_2r^4\\). Substituting the given values gives \\(1215=15r^4\\), and therefore \\(r^4=81\\). Since the ratio is stated to be positive, the suitable fourth root is \\(r=3\\), not \\(r=-3\\).

The seventh term is one more multiplication by the ratio after the sixth term. Thus \\(a_7=a_6r=1215\\times3=3645\\). Therefore option B is correct. The condition that the ratio is positive is important because the equation for \\(r^4\\) alone would also allow a negative fourth root, but that possibility is excluded by the question.

The first four terms are (3,12,48,192), and their product is (589824). Multiply each term carefully in product questions.

The general-term formula of a GP is \(a_n=a_1r^{n-1}\). Thus, \(a_4=10r^3=270\), so \(r^3=27\) and the positive value is \(r=3\). If 2 were used, the fourth term would be \(10\times2^3=80\), not 270. Exam tip: in \(a_n\), the exponent of \(r\) is always \(n-1\).

(\frac{a_7}{a_4}=r^3=4^3=64). In the same progression, the ratio of terms depends on the difference of positions.

In a geometric progression, moving two positions forward multiplies a term by r^2. Thus a_5 = a_3r^2, so 360 = 40r^2 and r^2 = 9. Since r is positive, r = 3. Moving from the fifth term to the seventh term also involves two steps, so a_7 = a_5r^2 = 360 × 9 = 3240. Therefore option C is correct. An alternative check uses a_7 = a_3r^4 = 40 × 3^4 = 40 × 81 = 3240. The other options do not result when the same two-step multiplier is applied consistently, and using r rather than r^2 would incorrectly ignore one of the two transitions.