Geometric Progression

(4\cdot4^{n-1}=4096) gives (4^n=4096=4^6), so (n=6). In exams, equate the last term to the general term.

The first four terms are 7, 21, 63, and 189. Therefore, \(S_4=7+21+63+189=280\). This can also be checked using \(S_n=\frac{a(r^n-1)}{r-1}\), where \(a=7\) and \(r=3\). The option 294 may result from an incorrect addition involving the last term. Exam tip: for a small number of terms, add directly and verify with the GP sum formula if needed.

From (2\cdot4^{n-1}=8192), (4^{n-1}=4096=4^6), so (n=7). In exams, equate powers to find the term number.

In a geometric progression, the nth term is given by \(a_n=a_1r^{n-1}\). Therefore, \(a_8=9(-2)^{8-1}=9(-2)^7=9(-128)=-1152\). Hence, option A is correct. \(1152\) results from a sign error, since an odd power of \((-2)\) is negative. Exam tip: When the common ratio is negative, first check whether the exponent is even or odd.

The sum of the first four terms is (5+30+180+1080=1295). In exams, direct addition is safe when the number of terms is small.

There are (5) steps from the fourth to the ninth term, so (a_9=162\cdot3^5=39366). In exams, count the position gap.

Each term is multiplied by (\frac{1}{3}), so the terms are (486,162,54,18,6,2). In exams, you can also check a decreasing GP in order.

From \(405\left(\frac{1}{3}\right)^{n-1}=5\), \(\left(\frac{1}{3}\right)^{n-1}=\frac{1}{81}\), so \(n=5\). In exams, simplify fractions first.

Here, the first term is \(a=10\), the common ratio is \(r=2\), and the number of terms is \(n=9\). The sum of the first \(n\) terms of a GP is \(S_n=\frac{a(r^n-1)}{r-1}\). Thus, \(S_9=\frac{10(2^9-1)}{2-1}=10(512-1)=5110\). Option 5120 is only \(10\times2^9\); it misses the required \(-1\) in the sum formula. Exam tip: before applying the formula, identify \(a\), \(r\), and \(n\).

In a GP, \(a_{n-1}=a_n/r\) and \(a_{n+1}=a_nr\). Multiplying gives \(a_{n-1}a_{n+1}=a_n^2\). Option D describes an arithmetic progression, where the common difference is constant. Exam tip: square the middle term.

With a positive first term, each next term is obtained by multiplying the previous term by a number between 0 and 1. Hence every next term is smaller, so the GP decreases. Exam tip: for 0<r<1, a positive GP is decreasing.

The terms are \(96,48,24,12,6,3\), and their sum is \(189\). In exams, direct addition is also easy for small decreasing terms.

From (\frac{2500}{20}=125=r^3), (r=5) and (a_1=20\div5=4). In exams, find (r) first and then (a_1).

From (15\cdot2^{n-1}=3840), (2^{n-1}=256=2^8), so (n=9). In exams, factor first and compare powers.

The direct answer is option B: \(a_5=810\). This is a geometric progression because each term is obtained by multiplying by the same positive ratio \(r\). The nth-term rule is \(a_n=a_1r^{n-1}\). For the fourth term, \(a_4=10r^3=270\). Dividing by 10 gives \(r^3=27\). Since \(r\) is positive, \(r=3\), not a negative cube root. The next term is obtained by multiplying the fourth term by 3: \(a_5=270\times3=810\). Equivalently, \(a_5=10\times3^4=810\). Option A, 540, would correspond to multiplying 270 by 2, but the established ratio is 3. Option B is correct because it follows from the ratio and the fourth term. Option C, 1080, would require a ratio of 4 after the fourth term, which is unsupported. Option D, 1350, would require a ratio of 5, also unsupported. The condition that \(r\) is positive matters because solving \(r^3=27\) gives the positive value 3. Exam cue: first find the ratio from the known terms, then multiply once more to get the next term.

Here, the first term is \(a=12\), the common ratio is \(r=2\), and \(n=10\). Using the GP sum formula \(S_n=\frac{a(r^n-1)}{r-1}\), we get \(S_{10}=\frac{12(2^{10}-1)}{2-1}=12(1024-1)=12276\). Hence, option A is correct. The close distractor 12288 equals \(12\times1024\) and results from omitting the \(-1\) term. Exam tip: remembering \(2^{10}=1024\) makes this calculation quicker.

In a geometric progression, the eighth term is \(a_8=ar^7\). Thus, \(1280=10r^7\), so \(r^7=128=2^7\). Since \(r\) is positive, \(r=2\). If \(r=3\), then \(10\times3^7\) would not equal 1280. Exam tip: in the \(n\)th term, the exponent of \(r\) is always \(n-1\).

Each term is multiplied by (\frac{1}{3}), and the sixth term is (\frac{40}{81}). In exams, fractional terms can also be checked in order.

(a_4=4\cdot6^3=864) and (a_5=5184), so the sum is (6048). In exams, calculate large powers separately.

(S_n=\frac{5(3^n-1)}{3-1}), and (\frac{5(3^n-1)}{2}=1820) gives (3^n=729), so (n=6). In exams, simplify the sum and identify the power.

For consecutive GP terms, \(\frac{y}{x}=\frac{z}{y}\). Cross-multiplying gives \(y^2=xz\), so A is correct. \(x+z=2y\) is the condition for an AP, not a GP. Exam tip: equate consecutive ratios first.