Geometric Progression

In a GP, the seventh term is \(a_7=ar^6\). Thus, \(1458=2r^6\), so \(r^6=729=3^6\). Since \(r\) is specified as positive, \(r=3\). For example, choosing 2 gives \(2\times2^6=128\), not 1458. Exam tip: in \(a_n=ar^{n-1}\), the exponent is always \(n-1\).

From (8\cdot4^{n-1}=8192), (4^{n-1}=1024=4^5), so (n=6). In exams, equate powers to find (n).

\(a_9=256\cdot\left(\frac{1}{2}\right)^8=1\). In exams, apply powers of fractional ratios carefully.

The governing concept is verification of a finite geometric-series sum. The first term is a = 12, the ratio is r = 4, and n = 4. Applying S_n = a(r^n - 1)/(r - 1), we obtain S_4 = 12(4^4 - 1)/(4 - 1) = 12(256 - 1)/3 = 12 × 255/3 = 4 × 255 = 1020. The same result follows by listing the terms: 12, 48, 192, and 768; their total is 1020. Therefore the claim that the sum is 780 is false, and option D gives the correct reason. Options A, B, and C do not equal the actual sum and therefore cannot correctly evaluate the statement.

The governing concept is the finite sum of a geometric progression. From the sequence, the first term is a = 5 and the common ratio is r = 20/5 = 4. With n = 6 and r greater than 1, use S_n = a(r^n - 1)/(r - 1). Thus S_6 = 5(4^6 - 1)/(4 - 1) = 5(4096 - 1)/3 = 5 × 4095/3 = 5 × 1365 = 6825. A direct check gives the terms 5, 20, 80, 320, 1280, and 5120, whose sum is 6825. Therefore option A is correct. The other values are close distractors caused by subtraction or arithmetic mistakes and do not satisfy the geometric-series formula.

(\frac{a_7}{a_3}=r^4=\frac{576}{36}=16), so (r=2). In exams, the ratio of distant terms gives a power based on the position gap.

In B, \(a_{n+1}^2=a_n a_{n+2}\) gives \(a_{n+1}/a_n=a_{n+2}/a_{n+1}\), so the common ratio is fixed. Option A indicates an AP, not a GP. Exam tip: test this product relation for consecutive triples.

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=6) and (r=4), so (S_5=\frac{6(4^5-1)}{4-1}=2046). In exams, identify (a) and (r) from the general term.

From (7\cdot3^{n-1}=5103), (3^{n-1}=729=3^6), so (n=7). In exams, compare powers to find the term number.

(\frac{a_6}{a_2}=r^4=\frac{5120}{20}=256), so (r=4). In exams, a gap of (4) positions gives (r^4).

Each term is multiplied by (\frac{1}{2}), and the terms are (640,320,160,80,40,20,10,5). In exams, you can also check a decreasing GP in order.

In a geometric progression, the sixth term is \(a_6=ar^{6-1}=ar^5\). Thus, \(416=13r^5\), so \(r^5=32=2^5\). Since \(r\) is positive, \(r=2\). If \(r=3\), then \(13\times3^5\) is not 416. Exam tip: for the \(n\)th term, use the exponent \(n-1\).

Here, the first term is \(a=11\), 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 \(S_n=\frac{a(r^n-1)}{r-1}\). Thus, \(S_6=\frac{11(3^6-1)}{3-1}=\frac{11(729-1)}{2}=4004\). Therefore, 4004 is correct. An option such as 3993 can result from an error while evaluating the power or subtraction. Exam tip: identify \(a\), \(r\), and \(n\) before substituting into the formula.

(\frac{5120}{80}=64=r^3), so (r=4) and (a_1=\frac{80}{4^2}=5). In exams, find (r) first and then the first term.

The fifth term is (x\cdot4^4=256x=1280), so (x=5). In exams, put the algebraic first term in (ar^{n-1}) too.

(3\cdot4^{n-1}=12288), so (4^{n-1}=4096=4^6) and (n=7). In exams, equate the last term to the general term.

Here, (S_4) denotes the sum of the first four terms: 5 + 25 + 125 + 625 = 780. Therefore, the correct answer is 780. The value 775 is only the sum of the first three terms, so it is a close but incorrect option. Exam tip: When there are only a few terms, direct addition is usually the quickest and safest method.

From (9\cdot4^{n-1}=9216), (4^{n-1}=1024=4^5), so (n=6). In exams, equate powers to find the term number.

The nth term of a geometric progression is \(a_n=a_1r^{n-1}\). Thus, \(a_7=6(-2)^{7-1}=6(-2)^6=6\times64=384\). Therefore, \(384\) is correct. \(-384\) would result if the power of \(-2\) were odd. Exam tip: With a negative common ratio, always check whether the exponent is even or odd.