Geometric Progression

There are (5) steps from the fourth to the ninth term, so (a_9=162\cdot3^5=39366). In exams, count the position gap.

Each term is multiplied by (\frac{1}{3}), so the terms are (486,162,54,18,6,2). In exams, you can also check a decreasing GP in order.

The relevant concept is the sum of the first four terms of a geometric progression. Since r = 4 is not 1, use Sₙ = a(rⁿ − 1)/(r − 1). With a = 8 and n = 4, S₄ = 8(4⁴ − 1)/(4 − 1) = 8(256 − 1)/3 = 8 × 255/3 = 8 × 85 = 680. A direct check gives the terms 8, 32, 128, and 512; adding them yields 8 + 32 + 128 + 512 = 680. Consequently, the stated value is true and option A is correct. Options B, C, and D are incorrect totals. They may reflect a wrong power, omission or alteration of a term, or a simple arithmetic error, but none agrees with both the formula and direct addition.

The governing concept is the general term of a geometric progression, aₙ = arⁿ⁻¹. The first term is a = 3 and the common ratio is r = 12/3 = 4. Therefore a₅ = 3 × 4⁴ = 3 × 256 = 768, while a₆ = 3 × 4⁵ = 3 × 1024 = 3072. Adding the two requested terms gives a₅ + a₆ = 768 + 3072 = 3840, so option B is correct. Option A is only the sixth term and does not include a₅. Options C and D can result from using an incorrect exponent or adding incorrectly. Computing both terms separately is the safest approach because the question asks for a sum of two terms, not for a single term or for the sum of all terms.

From \(405\left(\frac{1}{3}\right)^{n-1}=5\), \(\left(\frac{1}{3}\right)^{n-1}=\frac{1}{81}\), so \(n=5\). In exams, simplify fractions first.

Here, the first term is \(a=10\), the common ratio is \(r=2\), and the number of terms is \(n=9\). The sum of the first \(n\) terms of a GP is \(S_n=\frac{a(r^n-1)}{r-1}\). Thus, \(S_9=\frac{10(2^9-1)}{2-1}=10(512-1)=5110\). Option 5120 is only \(10\times2^9\); it misses the required \(-1\) in the sum formula. Exam tip: before applying the formula, identify \(a\), \(r\), and \(n\).

In a GP, \(a_{n-1}=a_n/r\) and \(a_{n+1}=a_nr\). Multiplying gives \(a_{n-1}a_{n+1}=a_n^2\). Option D describes an arithmetic progression, where the common difference is constant. Exam tip: square the middle term.

With a positive first term, each next term is obtained by multiplying the previous term by a number between 0 and 1. Hence every next term is smaller, so the GP decreases. Exam tip: for 0<r<1, a positive GP is decreasing.

The terms are \(96,48,24,12,6,3\), and their sum is \(189\). In exams, direct addition is also easy for small decreasing terms.

From (\frac{2500}{20}=125=r^3), (r=5) and (a_1=20\div5=4). In exams, find (r) first and then (a_1).

This is an explicit general rule for a geometric progression. Substitute n = 4 and n = 5 separately into aₙ = 6 × 5ⁿ⁻¹. We get a₄ = 6 × 5³ = 6 × 125 = 750 and a₅ = 6 × 5⁴ = 6 × 625 = 3750. Therefore a₄ + a₅ = 750 + 3750 = 4500. Option A is correct. Option B could result from an inaccurate addition, option C from using a wrong power or term value, and option D is another overestimate. The exponent n − 1 is important: the first term uses 5⁰, so replacing it with 5ⁿ would shift every term and produce an incorrect result.

From (15\cdot2^{n-1}=3840), (2^{n-1}=256=2^8), so (n=9). In exams, factor first and compare powers.

From (270=10r^3), (r^3=27) and (r=3), so (a_5=270\cdot3=810). In exams, find the ratio first.

Here, the first term is \(a=12\), the common ratio is \(r=2\), and \(n=10\). Using the GP sum formula \(S_n=\frac{a(r^n-1)}{r-1}\), we get \(S_{10}=\frac{12(2^{10}-1)}{2-1}=12(1024-1)=12276\). Hence, option A is correct. The close distractor 12288 equals \(12\times1024\) and results from omitting the \(-1\) term. Exam tip: remembering \(2^{10}=1024\) makes this calculation quicker.

In a geometric progression, the eighth term is \(a_8=ar^7\). Thus, \(1280=10r^7\), so \(r^7=128=2^7\). Since \(r\) is positive, \(r=2\). If \(r=3\), then \(10\times3^7\) would not equal 1280. Exam tip: in the \(n\)th term, the exponent of \(r\) is always \(n-1\).

Each term is multiplied by (\frac{1}{3}), and the sixth term is (\frac{40}{81}). In exams, fractional terms can also be checked in order.

(a_4=4\cdot6^3=864) and (a_5=5184), so the sum is (6048). In exams, calculate large powers separately.

(S_n=\frac{5(3^n-1)}{3-1}), and (\frac{5(3^n-1)}{2}=1820) gives (3^n=729), so (n=6). In exams, simplify the sum and identify the power.

For consecutive GP terms, \(\frac{y}{x}=\frac{z}{y}\). Cross-multiplying gives \(y^2=xz\), so A is correct. \(x+z=2y\) is the condition for an AP, not a GP. Exam tip: equate consecutive ratios first.

Direct answer: the sixth term is 2, so B is correct. Each term is half of the preceding term: 32 ÷ 64 = 1/2 and 16 ÷ 32 = 1/2. Thus the common ratio is r = 1/2. Counting carefully gives a₁ = 64, a₂ = 32, a₃ = 16, a₄ = 8, a₅ = 4, and a₆ = 2. The formula confirms the list: aₙ = arⁿ⁻¹, so a₆ = 64(1/2)⁵ = 64/32 = 2. D, 8, is the fourth term, and C, 4, is the fifth term. A, 1, would be the seventh term after one more division by 2. The common mistake is to count the number of divisions rather than the term number: reaching the sixth term requires five multiplications by 1/2 because the first term is already given. Memory cue: term number n means n−1 ratio steps from the first term.