Geometric Progression

(\frac{a_5}{a_2}=r^3=\frac{324}{12}=27), so (r=3). In exams a gap of (3) positions gives (r^3).

From \(729\left(\frac{1}{3}\right)^{n-1}=1\), \(3^{6-(n-1)}=1\), so \(n=7\). In exams count decreasing powers carefully.

For a geometric progression, the seventh term is \(a_7=ar^6\). Hence, \(192=3r^6\), so \(r^6=64=2^6\). Since \(r\) is stated to be positive, \(r=2\). If \(r=3\), then \(3\times3^6\) is not 192. Exam tip: in \(a_n=ar^{n-1}\), the exponent of \(r\) is always \(n-1\).

Here, the first term is \(a=4\), the common ratio is \(r=3\), and the number of terms is \(n=7\). The sum of the first \(n\) terms of a GP is \(S_n=\frac{a(r^n-1)}{r-1}\). Therefore, \(S_7=\frac{4(3^7-1)}{3-1}=\frac{4(2187-1)}{2}=4372\). Hence, 4372 is correct. An answer such as 4374 can result from a small error while evaluating \(3^7-1\) or dividing. Exam tip: write down \(a\), \(r\), and \(n\) separately before substituting in the formula.

(\frac{384}{48}=8=r^3), so (r=2) and (a_1=\frac{48}{2^2}=12). In exams find (r) first and then the first term.

The fifth term is (x\cdot2^4=16x=80), so (x=5). In exams put the algebraic first term in (ar^{n-1}) too.

The governing idea is the nth-term formula aₙ = arⁿ⁻¹. In this progression, a = 1 and r = 4. To find which term equals 1024, write 1 × 4ⁿ⁻¹ = 1024. Since 1024 = 4⁵, we get n − 1 = 5 and therefore n = 6. The terms confirm this: 1, 4, 16, 64, 256, 1024, so there are six terms through 1024. Hence option B is correct. Option A would stop at 256, the fifth term. Option C would include the next term, 4096, and option D would go even farther. The phrase “up to 1024” means the count ends when 1024 is reached.

Here, \(S_4\) is the sum of the first four terms: \(2+8+32+128=170\). Therefore, the correct answer is 170. Option 168 is incorrect because the sum of all four given terms is 170. Exam tip: when only a few terms are given, direct addition is often the quickest and safest method.

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.

The governing concept is the finite sum of a geometric progression. The first five terms can be generated by repeatedly multiplying by r = 1/3: 81, 27, 9, 3 and 1. Adding them gives 81 + 27 + 9 + 3 + 1 = 121, so option B is correct. The formula confirms this result: S₅ = a(1 − r⁵)/(1 − r) = 81[1 − (1/3)⁵]/(1 − 1/3) = 121. Because the ratio is less than one, the terms decrease, but all five terms must still be included. Option A omits one unit, while options C and D are larger than the correctly calculated sum. Direct addition is especially efficient here because the number of terms is small.