Geometric Progression

From (6\cdot4^{n-1}=6144), (4^{n-1}=1024=4^5), so (n=6). In exams equate powers to find the term number.

The outputs are 160, 240, 360 and 540, so \(160+240+360+540=1300\). The total 1120 assumes a fixed increase of 160, which is arithmetic progression reasoning. Exam tip: verify the common ratio first.

Here, the first term is \(a=3\), the common ratio is \(r=5\), and \(n=5\). Therefore, \(S_5=\frac{a(r^5-1)}{r-1}=\frac{3(5^5-1)}{5-1}=2343\). Hence, 2343 is correct. A value such as 2340 can result from a minor arithmetic error while adding or applying the formula. Exam tip: identify \(a\), \(r\), and \(n\) before using the GP sum formula.

There are (5) steps from the fourth to the ninth term, so (a_9=80\cdot2^5=2560). In exams count the position gap.

From \(512\left(\frac{1}{2}\right)^{n-1}=4\), \(\left(\frac{1}{2}\right)^{n-1}=\frac{1}{128}\), so \(n=8\). In exams count terms while halving.

For a geometric progression, the sum of the first n terms is \(S_n=\frac{a(r^n-1)}{r-1}\), where \(r\ne1\). Here, \(S_4=\frac{6(3^4-1)}{3-1}=\frac{6(81-1)}{2}=240\). Therefore, the given statement is true. The values 120, 180, and 300 are not the sum of the first four terms of this GP. Exam tip: while finding a sum, use the number of terms as the exponent in \(r^n\).

The governing concept is the nth term of a geometric progression. The first term is a = 8 and the common ratio is r = 24 ÷ 8 = 3, so aₙ = 8 × 3ⁿ⁻¹. Hence a₅ = 8 × 3⁴ = 8 × 81 = 648, and a₆ = 8 × 3⁵ = 8 × 243 = 1944. Their sum is 648 + 1944 = 2592. Therefore the corrected and unique answer is option C. Option A is only half of the total, while option B can arise from using an incorrect power or adding inaccurately. Option D is not the required sum either. The original option set listed 1080 for option C, but that value is inconsistent with the progression; replacing it with 2592 makes the question mathematically valid.

From \(100\left(\frac{1}{2}\right)^{n-1}=\frac{25}{2}\), \(\left(\frac{1}{2}\right)^{n-1}=\frac{1}{8}\), so \(n=4\). In exams simplify fractions first.

For this GP, the first term is \(a=4\), the common ratio is \(r=2\), and \(n=10\). Using \(S_n=\frac{a(r^n-1)}{r-1}\), \(S_{10}=\frac{4(2^{10}-1)}{2-1}=4(1024-1)=4092\). Hence, option A is correct. \(4096\) is only \(4\times2^{10}\); it misses the \(-1\) part of the sum formula. Exam tip: when \(r\ne1\), do not forget the denominator \(r-1\) in the GP sum formula.

In a GP, \(\frac{y}{x}=\frac{z}{y}\). Cross-multiplying gives \(y^2=xz\), so option A is correct. \(x+z=2y\) is the condition for an AP, not a GP. Exam tip: square the middle term and compare it with the product of the extremes.

Dividing the condition by \(a_{n-1}a_n\) gives \(\frac{a_n}{a_{n-1}}=\frac{a_{n+1}}{a_n}\). Thus, consecutive terms have a constant ratio, so it is a GP. Exam tip: check the non-zero condition before dividing.

From (\frac{486}{18}=27=r^3), (r=3) and (a_1=18\div3=6). In exams find (r) first and then (a_1).

From (11\cdot2^{n-1}=1408), (2^{n-1}=128=2^7), so (n=8). In exams factor first and compare powers.

In a GP, consecutive ratios are equal. From \(q^2=pr\), \(q/p=r/q\); from \(r^2=qs\), \(r/q=s/r\). Hence all ratios are equal. Exam tip: divide only after confirming that the terms are non-zero.

Here, the first term is \(a=9\), the common ratio is \(r=2\), and \(n=12\). The sum of the first \(n\) terms of a GP is \(S_n=\frac{a(r^n-1)}{r-1}\). Thus, \(S_{12}=\frac{9(2^{12}-1)}{2-1}=9(4096-1)=36855\). The option 36864 equals \(9\times4096\); it results from missing the \(-1\). Exam tip: when \(r\ne1\), remember the \(r^n-1\) part of the sum formula.

In a geometric progression, the sixth term is \(a_6=ar^5\). Thus, \(224=7r^5\), so \(r^5=32=2^5\). Since \(r\) is positive, \(r=2\). If \(r=3\), then \(7\times3^5\) is obtained, not 224. Exam tip: in \(a_n=ar^{n-1}\), the exponent of \(r\) is always \(n-1\).

Each term is multiplied by (\frac{1}{3}), and the sixth term is (\frac{5}{27}). In exams fractional terms can also be checked in order.

(a_5=2\cdot4^4=512) and (a_6=2048), so the sum is (2560). In exams calculate large powers separately.

(S_n=2(3^n-1)/(3-1)=3^n-1), and (3^n-1=242) gives (n=5). In exams simplify the sum and identify the power.

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}).

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.