Geometric Progression

The direct answer is Option B: the sixth term. The first term is 8, and the common ratio is 32/8=4. Therefore the nth term is
(a_n=8\cdot4^{n-1}). Set this equal to 8192:
(8\cdot4^{n-1}=8192). Divide by 8 to get
(4^{n-1}=1024). Since
(1024=4^5), we have n-1=5, so n=6. A direct listing confirms it: 8, 32, 128, 512, 2048, 8192. Option A, the fifth term, is 2048, not 8192. Option B is correct because 8192 appears after multiplying 2048 by 4. Option C, the seventh term, would be 32768, so it is one term too late. Option D, the eighth term, would be 131072 and is even farther away. The common mistake is to count the exponent as the term number; the exponent is n-1 because the first term has exponent zero. Exam cue: in a GP, write
(a_n=ar^{n-1}), compare powers, then add one to the exponent.

\(a_9=256\cdot\left(\frac{1}{2}\right)^8=1\). 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.

The direct answer is Option A: 777. The common ratio is 18/3=6, and each following term is six times the previous one. The first four terms are already given: 3, 18, 108 and 648. Add them carefully: 3+18=21, 21+108=129, and 129+648=777. Therefore the required sum is 777. Option A is correct because it exactly equals the sum of all four requested terms. Option B, 780, is 3 more than the actual total and may result from an addition error. Option C, 783, is 6 more than the actual total and has no valid basis. Option D, 786, is 9 more than the actual total and is also incorrect. The geometric-sum formula would give the same result:
(S_4=a(r^4-1)/(r-1)=3(6^4-1)/5=777), but direct addition is safer here because only four terms are involved. Remember to include every term exactly once; do not continue the sequence or add the ratio instead of adding the terms.

There are (6) steps from the fourth to the tenth term, so (a_{10}=96\cdot2^6=6144). In exams, count the position gap.

Each term is multiplied by (\frac{1}{3}), so the terms are (810,270,90,30,10). In exams, you can also check a decreasing GP in order.

The direct answer is option A: the statement is true because \(S_5=1210\). In a geometric progression, each term is obtained by multiplying the previous term by the common ratio. The sum of the first \(n\) terms, when \(r\ne1\), is \(S_n=\frac{a(r^n-1)}{r-1}\). Here \(a=10\), \(r=3\), and \(n=5\). Therefore, \(S_5=\frac{10(3^5-1)}{3-1}=\frac{10(243-1)}{2}=\frac{10\times242}{2}=1210\). Option A is correct because its value agrees exactly with the calculation. Option B says 1200, but this does not result from the formula. Option C says 1220, which is also an incorrect arithmetic result. Option D says 2430; this is not the sum of five terms and is closer to an unrelated multiplication involving powers. A useful check is to list the terms: \(10,30,90,270,810\); their sum is \(1210\). Memory cue: for a GP sum, substitute carefully and check by adding the terms when the numbers are manageable.

The governing concept is the general term of a geometric progression. The first term is a = 4 and the common ratio is r = 16/4 = 4. The nth term is a_n = ar^(n-1). Therefore a_5 = 4 × 4^4 = 4 × 256 = 1024, and a_6 = 4 × 4^5 = 4 × 1024 = 4096. Adding these two required terms gives a_5 + a_6 = 1024 + 4096 = 5120. Hence option B is correct. Option A is only a_6, so it omits a_5. Options C and D are larger sums that do not follow from the specified two terms. Finding each requested term before adding prevents confusing a term with the sum of terms.

From \(162\left(\frac{1}{3}\right)^{n-1}=2\), \(\left(\frac{1}{3}\right)^{n-1}=\frac{1}{81}\), so \(n=5\). In exams, simplify fractions first.