Geometric Progression

Consecutive GP terms have equal ratios, so \(y/x=z/y\). Cross-multiplication gives \(y^2=xz\). The relation \(x+z=2y\) belongs to an AP. Exam tip: always preserve the stated order of terms.

Here \(r=\frac{1}{5}\), so the sixth term is \(625\cdot\left(\frac{1}{5}\right)^5=\frac{1}{5}\). In exams, apply fractional ratios carefully in decreasing GPs.

From (6\cdot2^{n-1}=1536), (2^{n-1}=256=2^8), so (n=9). In exams, equate powers to find the term number.

In a GP, the seventh term is \(a_7=ar^6\). Thus, \(320=5r^6\), so \(r^6=64=2^6\). Since \(r\) is stated to be positive, \(r=2\); although \(-2\) also has sixth power 64, it is not positive. Exam tip: the exponent of \(r\) in the \(n\)th term is \(n-1\).

The first term is (12) and the ratio is (4), so (a_n=12\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. The first term is a = 3, the common ratio is r = 3, and six terms are required. Because r ≠ 1, use Sₙ = a(rⁿ − 1)/(r − 1). Substitution gives S₆ = 3(3⁶ − 1)/(3 − 1) = 3(729 − 1)/2 = 3 × 728/2 = 1092. A direct verification lists the terms as 3, 9, 27, 81, 243, and 729; adding them gives 3 + 9 + 27 + 81 + 243 + 729 = 1092. Therefore option A is correct. The nearby alternatives 1093, 1095, and 1098 are not valid sums and could result from an addition slip, an incorrect power, or an arbitrary adjustment.

For the middle term, (x^2=12\cdot75=900), so (x=30). In exams, the square of the middle GP term equals the product of the outer terms.

The first five terms are (4,20,100,500,2500), and the sum is (3124). In exams, add terms carefully when the ratio is large.

In a GP, the ratio of consecutive terms remains fixed: \(a_{n+1}/a_n=r\). Option B describes a constant difference, which identifies an AP. Exam tip: check ratios, not differences, to identify a GP.

The first term is (100) and the ratio is \(\frac{1}{2}\), so the correct rule is \(100\cdot\left(\frac{1}{2}\right)^{n-1}\). In exams, write the fractional ratio in a decreasing GP.

The nth term of a GP is \(a_n=ar^{n-1}\). Thus, \(a_5=9r^4=2304\), so \(r^4=256=4^4\). Since \(r\) is stated to be positive, \(r=4\). If \(r=2\), then \(9\times2^4=144\), not 2304. Exam tip: for the fifth term, use the exponent \(5-1=4\).

From (6\cdot3^{n-1}=4374), (3^{n-1}=729=3^6), so (n=7). In exams, equate powers to find (n).

\(a_8=192\cdot\left(\frac{1}{2}\right)^7=\frac{3}{2}\). In exams, apply powers of fractional ratios carefully.

The governing concept is the finite sum of the first n terms of a geometric progression. Here a = 4, r = 4, and n = 4. Since r ≠ 1, Sₙ = a(rⁿ − 1)/(r − 1). Thus S₄ = 4(4⁴ − 1)/(4 − 1) = 4(256 − 1)/3 = 4 × 255/3 = 340. Direct listing gives the same result: the four terms are 4, 16, 64, and 256, and their sum is 4 + 16 + 64 + 256 = 340. Therefore the stated value is true and option A is correct. Option D confuses the final term, 256, with the sum. Options B and C agree with neither the formula nor direct addition.

The governing concept is the sum formula for a geometric progression. From 2, 8, 32 and 128, the first term is a = 2 and the common ratio is r = 4. For seven terms, use S_n = a(r^n − 1)/(r − 1). Hence S_7 = 2(4^7 − 1)/(4 − 1). Since 4^7 = 16,384, this becomes 2(16,383)/3 = 32,766/3 = 10,922. Therefore option B is correct. Direct addition of 2 + 8 + 32 + 128 + 512 + 2048 + 8192 also gives 10,922. The adjacent alternatives differ by small amounts and reflect arithmetic or power-calculation errors, not a valid geometric sum.

(\frac{a_6}{a_2}=r^4=\frac{224}{14}=16), so (r=2). In exams, the ratio of distant terms gives a power based on the position gap.

In a GP, \(y=xr\) and \(z=xr^2\). Hence \(y^2=(xr)^2=x(xr^2)=xz\). The relation \(x+y=2z\) is associated with an AP, not a GP. Exam tip: square the middle term and compare it with the product of the outer terms.

The position gap is (5) and (r=2), so the ratio is (2^5=32). In exams, use (\frac{a_m}{a_n}=r^{m-n}).

Here (a=5) and (r=3), so (S_6=\frac{5(3^6-1)}{3-1}=1820). In exams, identify (a) and (r) from the general term.

From (4\cdot3^{n-1}=972), (3^{n-1}=243=3^5), so (n=6). In exams, compare powers to find the term number.