Geometric Progression

(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.

The direct answer is option C: \(a_5=2500\). For a geometric progression, \(a_n=a_1r^{n-1}\). The fourth term is three ratio steps after the first, so \(a_4=4r^3\). Using \(a_4=500\), we get \(500=4r^3\), hence \(r^3=125\). Because the ratio is positive, \(r=5\). The next term is therefore \(a_5=a_4r=500\times5=2500\). Option A, 1500, does not use the ratio obtained from the given terms. Option B, 2000, is also not the result of multiplying 500 by 5. Option C is correct because it follows directly from the verified ratio. Option D, 3000, would require a ratio of 6, which would make the fourth term inconsistent with the data. A quick check is the sequence \(4,20,100,500,2500\), where every term is five times the previous one. Memory cue: when the first and fourth terms are known, solve for \(r^3\), then multiply the fourth term by \(r\).

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}).

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.

The \(n\)th term of a GP is \(a_n=ar^{n-1}\). Thus, \(a_6=3r^5=96\), so \(r^5=32=2^5\). Since \(r\) is positive, \(r=2\). If \(r=3\), then \(3\times3^5\) is not equal to 96. Exam tip: for the sixth term, the exponent of \(r\) is \(6-1=5\).

The direct answer is option B: 5120 is the sixth term. In a geometric progression, each term is obtained by multiplying the previous term by the same number, called the common ratio. Here, the terms are 5, 20, 80, 320, so the ratio is 4. The nth-term formula is \(a_n=a_1r^{n-1}\). Therefore \(5120=5\times4^{n-1}\). Dividing by 5 gives \(1024=4^{n-1}\). Since \(1024=4^5\), we get \(n-1=5\), so \(n=6\). A direct check also helps: first 5, second 20, third 80, fourth 320, fifth 1280, sixth 5120. Option A is wrong because the fifth term is 1280. Option B is correct because the sixth term is 5120. Option C is wrong because the seventh term would be \(5120\times4=20480\). Option D is wrong because the eighth term would be \(81920\). The important idea is that the exponent in a GP nth-term formula is one less than the term number. Memory cue: count the first term as exponent zero, not exponent one.

\(a_6=160\cdot\left(\frac{1}{2}\right)^5=5\). In exams, apply powers of fractional ratios carefully.

(\frac{a_8}{a_4}=r^4=\frac{768}{48}=16), so (r=2). In exams, the ratio of distant terms gives a power based on the position gap.

In a GP, the ratio of every term to its preceding term must remain constant. In option B, \(10/5=2\) and \(20/10=2\), but \(35/20=7/4\). Hence it is not a GP. Exam tip: check all consecutive ratios.

The position gap is (4) and (r=3), so the ratio is (3^4=81). In exams, use (\frac{a_m}{a_n}=r^{m-n}).

Here (a=9) and (r=2), so (S_8=9(2^8-1)=2295). In exams, identify (a) and (r) from the general term.

From (3\cdot5^{n-1}=9375), (5^{n-1}=3125=5^5), so (n=6). In exams, compare powers to find the term number.

(\frac{a_6}{a_2}=r^4=\frac{1458}{18}=81), so (r=3). In exams, a gap of (4) positions gives (r^4).

In a geometric progression, the seventh term is \(a_7=ar^6\). Thus, \(384=6r^6\), so \(r^6=64=2^6\). Since \(r\) is stated to be positive, \(r=2\). Although \(r=-2\) also satisfies \(r^6=64\), it violates the given condition. Exam tip: in \(a_n=ar^{n-1}\), the exponent of \(r\) is always \(n-1\).

Here, the first term is \(a=6\), 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, \(S_7=\frac{6(3^7-1)}{3-1}=\frac{6(2187-1)}{2}=6558\). Hence, option A is correct. A nearby value such as \(6560\) can result from an arithmetic error in the final calculation. Exam tip: write down \(a\), \(r\), and \(n\) separately before applying the formula.

(\frac{6250}{50}=125=r^3), so (r=5) and (a_1=\frac{50}{5^2}=2). In exams, find (r) first and then the first term.

The fifth term is (x\cdot5^4=625x=3125), so (x=5). In exams, put the algebraic first term in (ar^{n-1}) too.