Geometric Progression

In a geometric progression, the ratios of consecutive terms are equal. Here, the common ratio is \(\frac{96}{24}=4\), so \(\frac{24}{x}=4\). Hence, \(x=\frac{24}{4}=6\). Equivalently, for three consecutive GP terms, the square of the middle term equals the product of the outer terms: \(24^2=96x\), giving \(x=6\). If \(x=8\), the ratios are \(24/8=3\) and \(96/24=4\), which are not equal. Exam tip: For consecutive GP terms \(a,b,c\), use \(b^2=ac\) directly.

The first (5) terms are (36,18,9,\frac{9}{2},\frac{9}{4}), and their sum is (\frac{279}{4}). Add fractional terms using a common denominator.

For a geometric progression, the nth term is
. Hence,
and
. Therefore,
. The nearby option 660 does not result when the correct powers of the common ratio are used. Exam tip: for the nth term, the exponent of
is always
, not
.

In a geometric progression, the \(n\)th term is \(a_n=a_1r^{n-1}\). Therefore, \(a_4=27\left(\frac{2}{3}\right)^{4-1}=27\left(\frac{2}{3}\right)^3=27\times\frac{8}{27}=8\). The value 6 can result from incorrectly multiplying by the common ratio only twice. Exam tip: in \(a_n\), the exponent of the common ratio is always \(n-1\).

The common ratio is found by dividing consecutive terms: r = -12/3 = -4, and the later ratios confirm it. The nth-term formula is a_n = a_1r^(n-1). Thus a_6 = 3(-4)^5. Because the exponent 5 is odd, (-4)^5 = -1024, and multiplying by 3 gives a_6 = -3072. Therefore option D is correct. The sign alternates because the ratio is negative; after the fifth multiplication the result remains negative. Options A and C have positive signs, and option B has the wrong magnitude, usually caused by using one fewer factor of -4.

(a_1=\frac{22}{3}) and (a_4=198), so the sum is (\frac{616}{3}). The correct option should include (\frac{616}{3}).

(x^2=36\cdot100=3600), so positive (x=60). The square of the middle term equals the product of the outer terms.

(500=4r^3), so (r^3=125) and (r=5), hence (a_5=2500). First find the ratio and then form the next term.

The governing concept is the general-term rule aₙ = a₁rⁿ⁻¹. With r = 2, the second term is a₂ = a₁ × 2¹ = 2a₁, while the fourth term is a₄ = a₁ × 2³ = 8a₁. The exponent for the fourth term is 3 because three ratio multiplications take us from the first term to the fourth. Using the given relation, 2a₁ + 8a₁ = 170, so 10a₁ = 170 and a₁ = 17. Therefore option C is correct. Checking the alternatives, 12 would produce a sum of 120, 15 would produce 150, and 20 would produce 200. The key error to avoid is replacing a₄ = a₁r³ with 4a₁, which would not represent a geometric progression.

The governing concept is the nth-term formula for a geometric progression: aₙ = arⁿ⁻¹. In the given sequence, the first term is a = 2 and the common ratio is r = 6 ÷ 2 = 3; the same ratio continues because 18 ÷ 6 and 54 ÷ 18 are also 3. For the tenth term, substitute n = 10: a₁₀ = 2 × 3⁹. Since 3⁹ = 19683, the result is 2 × 19683 = 39366. Therefore option B is correct. Option A is only 3⁹ and omits the first-term factor 2; option C corresponds to 3¹⁰, while option D doubles the correct value again. The exponent is n − 1 because the first term has zero ratio multiplications.

The nth term of a geometric progression is \(a_n=ar^{n-1}\). Therefore, \(a_6=5\times4^{6-1}=5\times4^5=5\times1024=5120\). Hence, option C is correct. \(2560\) results from an error in evaluating the power of 4 or in multiplication. Exam tip: the exponent in the nth term is always \(n-1\), not \(n\).

Here \(r=\frac{1}{2}\), so \(a_9=1024\cdot\left(\frac{1}{2}\right)^8=4\). In exams, apply fractional ratios carefully in decreasing GPs.

From (7\cdot2^{n-1}=1792), (2^{n-1}=256=2^8), so (n=9). In exams, equate powers to find the term number.

The nth term of a geometric progression is aₙ = arⁿ⁻¹. Hence, for the fifth term, 4r⁴ = 324. So, r⁴ = 81 = 3⁴. Since r is stated to be positive, r = 3. Although r = −3 also gives r⁴ = 81, it is not positive. Exam tip: in the fifth term, the power of r is 4, not 5.

The first term is (6) and the ratio is (4), so (a_n=6\cdot4^{n-1}). In exams, keep (a) and (r) correct in (ar^{n-1}).

The governing concept is the finite sum of a geometric progression. For first term a and ratio r not equal to 1, use Sₙ = a(rⁿ − 1)/(r − 1) when r is greater than 1. Here a = 3, r = 3 and n = 7, so S₇ = 3(3⁷ − 1)/(3 − 1). Because 3⁷ = 2187, the numerator becomes 3(2186) = 6558, and division by 2 gives S₇ = 3279. Hence option B is correct. Directly adding the terms gives 3 + 9 + 27 + 81 + 243 + 729 + 2187 = 3279, which confirms the formula. The nearby alternatives result from arithmetic or endpoint errors; the sum must include exactly seven terms.

For the middle term, (x^2=12\cdot108=1296), so (x=36). In exams, the square of the middle GP term equals the product of the outer terms.

Here, the first term is \(a=8\), the common ratio is \(r=2\), and the number of terms is \(n=8\). The sum of the first \(n\) terms of a geometric progression is \(S_n=\frac{a(r^n-1)}{r-1}\). Thus, \(S_8=\frac{8(2^8-1)}{2-1}=8(256-1)=2040\). Therefore, option C is correct. Option 4080 incorrectly doubles the sum. Exam tip: identify \(a\), \(r\), and \(n\) before applying the GP sum formula.

In a GP, \(a_{n-1}=a_n/r\) and \(a_{n+1}=a_nr\). Multiplying gives \(a_{n-1}a_{n+1}=a_n^2\). Options B and D describe the equal-difference property of an AP. Exam tip: square the middle term to test a GP quickly.

The first term is (243) and the ratio is \(\frac{1}{3}\), so the correct rule is \(243\cdot\left(\frac{1}{3}\right)^{n-1}\). In exams, write the fractional ratio in a decreasing GP.