Geometric Progression

(\frac{a_7}{a_3}=r^4=\frac{1458}{18}=81), so (r=3). In exams, a gap of (4) positions gives (r^4).

From \(1024\left(\frac{1}{2}\right)^{n-1}=2\), \(2^{10-(n-1)}=2^1\), so \(n=10\). In exams, count decreasing powers carefully.

In a GP, the sixth term is \(a_6=ar^{6-1}=ar^5\). Thus, \(352=11r^5\), so \(r^5=32=2^5\). Since \(r\) is positive, \(r=2\). Taking \(r=3\) would give \(11\times3^5\), not 352. Exam tip: in \(a_n=ar^{n-1}\), remember that the exponent is \(n-1\).

Here, the first term is \(a=9\), the common ratio is \(r=3\), and the number of terms is \(n=6\). The sum of the first \(n\) terms of a GP is \(S_n=\frac{a(r^n-1)}{r-1}\). Therefore, \(S_6=\frac{9(3^6-1)}{3-1}=\frac{9(729-1)}{2}=3276\). Hence, option A is correct. Exam tip: identify \(a\), \(r\), and \(n\) before substituting into the formula.

(\frac{1024}{16}=64=r^3), so (r=4) and (a_1=\frac{16}{4}=4). In exams, find (r) first and then the first term.

The sixth term is (x\cdot3^5=243x=2430), so (x=10). In exams, put the algebraic first term in (ar^{n-1}) too.

(2\cdot4^{n-1}=2048), so (4^{n-1}=1024=4^5) and (n=6). In exams, equate the last term to the general term.

Here, \(S_4\) means the sum of the first four terms: \(3+15+75+375=468\). Therefore, 468 is correct. Note that 375 is the last term, not the sum; similarly, 475 is not obtained from the required addition. Exam tip: when only a few terms are given, direct addition is often the quickest way to verify the sum.

From (10\cdot4^{n-1}=10240), (4^{n-1}=1024=4^5), so (n=6). In exams, equate powers to find the term number.

In a geometric progression, \(a_n=a_1r^{n-1}\). Therefore, \(a_5=4(-3)^{5-1}=4(-3)^4=4\times81=324\). Hence, \(324\) is correct. \(-324\) would result only if the power of \(-3\) were odd. Exam tip: with a negative common ratio, first check whether the exponent is even or odd.

The sum of the first four terms is (2+14+98+686=800). In exams, direct addition is safe when the number of terms is small.

There are (6) steps from the third to the ninth term, so (a_9=72\cdot2^6=4608). In exams, count the position gap.

Each term is multiplied by (\frac{1}{3}), so the terms are (405,135,45,15,5). In exams, you can also check a decreasing GP in order.

For a geometric progression with \(r\ne1\), the sum of the first \(n\) terms is \(S_n=\frac{a(r^n-1)}{r-1}\). Thus, \(S_4=\frac{8(3^4-1)}{3-1}=\frac{8(81-1)}{2}=320\). Hence, the given statement is true. Values such as \(312\) or \(324\) result from an incorrect calculation. Exam tip: substitute \(n=4\) correctly in \(r^n\) before simplifying.

The governing concept is the general term of a geometric progression. In 2, 6, 18, the first term is a = 2 and the common ratio is r = 3. Using a_n = a × r^(n−1), we obtain a_6 = 2 × 3^5 = 2 × 243 = 486 and a_7 = 2 × 3^6 = 2 × 729 = 1458. Their sum is a_6 + a_7 = 486 + 1458 = 1944. Thus option A is correct. Option B may result from using an incorrect power, while C and D are larger values caused by extending the progression or adding terms incorrectly. Finding each requested term before adding avoids confusion between a term and a sum.

From \(128\left(\frac{1}{2}\right)^{n-1}=8\), \(\left(\frac{1}{2}\right)^{n-1}=\frac{1}{16}\), so \(n=5\). In exams, simplify fractions first.

Here, the first term is \(a=5\), the common ratio is \(r=2\), and the number of terms is \(n=10\). The sum of a geometric progression is \(S_n=\frac{a(r^n-1)}{r-1}\). Thus, \(S_{10}=\frac{5(2^{10}-1)}{2-1}=5(1024-1)=5115\). Therefore, option A is correct. Getting 5120 results from using \(2^{10}\) instead of \(2^{10}-1\). Exam tip: identify \(a\), \(r\), and \(n\) before applying the formula.

In a GP,
\(r=\frac{a_{n+1}}{a_n}\) remains constant for every consecutive pair. A constant difference instead identifies an AP. Exam tip: compare ratios, not differences.

In a GP, consecutive ratios are equal: \(\frac{b}{a}=\frac{c}{b}\). Cross-multiplication gives \(b^2=ac\). The condition \(2b=a+c\) belongs to an AP. Exam tip: square the middle term and compare it with the product of the outer terms.

The terms are (162,54,18,6,2), and the sum is (242). In exams, direct addition is also easy for small decreasing terms.