Geometric Progression

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 direct answer is Option B: 240. The first term is a=162 and the common ratio is r=
(1/3), so multiply each term by one third: 162, 54, 18, 6, 2. Add the first five terms step by step: 162+54=216, 216+18=234, 234+6=240, and 240+2=242. Therefore the sum is 242, not 240. The same result follows from the geometric-sum formula
(S_n=a(1-r^n)/(1-r)) when r is not 1:
(S_5=162(1-(1/3)^5)/(1-1/3)=242). Thus the supplied answer B is correct. Option A, 240, is the partial total after the fourth term and forgets the fifth term 2. Option B is correct because it includes all five terms. Option C, 244, adds two too much and does not match the sequence. Option D, 246, adds four too much and is also unsupported. A useful check is to write every requested term before adding; for a decreasing positive GP, the sum must be greater than the first four-term sum by exactly the fifth term.

From (\frac{1215}{45}=27=r^3), (r=3) and (a_1=45\div3^2=5). In exams, find (r) first and then (a_1).

From (12\cdot2^{n-1}=1536), (2^{n-1}=128=2^7), so (n=8). In exams, factor first and compare powers.

The direct answer is C, 3750. In a geometric progression, each term is obtained by multiplying the preceding term by the same common ratio. The formula is (a_n=a_1r^{n-1}). Here, (a_1=6) and (a_4=750), so (750=6r^3) because the fourth term has three multiplications by (r). Dividing by 6 gives (r^3=125). Since (r) is positive, (r=5). Therefore (a_5=a_4r=750\times5=3750). Option A, 2750, does not result from multiplying the fourth term by the correct ratio. Option B, 3250, is also not obtained from the progression. Option C is correct because it is the fifth term after multiplying by 5. Option D, 4500, would require a ratio of 6 from the fourth to the fifth term, which is inconsistent. Memory cue: in a GP, move one term forward by multiplying once by the common ratio.

For this GP, the first term is \(a=15\), the common ratio is \(r=2\), and \(n=9\). Using \(S_n=\frac{a(r^n-1)}{r-1}\), we get \(S_9=\frac{15(2^9-1)}{2-1}=15(512-1)=7665\). Hence, 7665 is correct. A value such as 7680 can result from an error in subtraction or in the final calculation. Exam tip: identify \(a\), \(r\), and \(n\) before substituting in the GP sum formula.

In a geometric progression, the seventh term is \(a_7=ar^6\). Thus, \(320=5r^6\), so \(r^6=64=2^6\). Since \(r\) is stated to be positive, \(r=2\). Although \(r=-2\) also has sixth power 64, it is excluded by the positive condition. Exam tip: in the \(n\)th term, the exponent of \(r\) is always \(n-1\).

Each term is multiplied by (\frac{1}{2}), and the seventh term is (\frac{3}{2}). In exams, fractional terms can also be checked in order.

(a_5=3\cdot5^4=1875) and (a_6=9375), so the sum is (11250). In exams, calculate large powers separately.

(S_n=\frac{4(3^n-1)}{3-1}=2(3^n-1)), and (2(3^n-1)=484) gives (n=5). In exams, simplify the sum and identify the power.

The sequence is geometric because each term is obtained by multiplying the previous term by 3. Its first term is \(a=5\), its common ratio is \(r=3\), and the nth-term formula is \(a_n=ar^{n-1}\). The exponent is one less than the term number because the first term has no multiplication by the ratio.

For the thirteenth term, \(a_{13}=5\cdot3^{12}\). Since \(3^{12}=531441\), multiplication by 5 gives \(2657205\). Therefore, option C is correct. A common error is to use \(3^{13}\), which would count one extra multiplication and produce a value that is too large.

The consecutive terms can be written as \(x, xr, xr^2\). Hence \(y^2=(xr)^2=x(xr^2)=xz\). The relation \(x+z=2y\) belongs to an arithmetic progression. Exam tip: square the middle term.

Here \(r=\frac{1}{3}\), so \(a_8=729\cdot\left(\frac{1}{3}\right)^7=\frac{1}{3}\). In exams, apply fractional ratios carefully in decreasing GPs.

From (8\cdot2^{n-1}=4096), (2^{n-1}=512=2^9), so (n=10). In exams, equate powers to find the term number.

For a geometric progression, the fifth term is \(a_5=ar^4\). Thus, \(2304=9r^4\), so \(r^4=256=4^4\). Since \(r\) is positive, \(r=4\). Option 3 is not correct because \(9\times3^4=729\), not 2304. Exam tip: use \(a_n=ar^{n-1}\) and check the exponent carefully.

The first term is (14) and the ratio is (3), so (a_n=14\cdot3^{n-1}). In exams, keep (a) and (r) correct in (ar^{n-1}).

The governing concept is the finite geometric-series sum. The first term is a = 4, and the common ratio is r = 12/4 = 3. For n = 6 and r ≠ 1, apply Sₙ = a(rⁿ − 1)/(r − 1). Therefore S₆ = 4(3⁶ − 1)/(3 − 1) = 4(729 − 1)/2 = 4 × 728/2 = 1456. A direct check produces the six terms 4, 12, 36, 108, 324, and 972, whose total is 4 + 12 + 36 + 108 + 324 + 972 = 1456. Thus option A is correct. Option B is a nearby arithmetic error, while C and D do not follow from the formula or from adding the listed six terms. Identifying the ratio before applying the formula is essential.

For the middle term, (x^2=18\cdot200=3600), so (x=60). In exams, the square of the middle GP term equals the product of the outer terms.

The first five terms are (6,30,150,750,3750), and their sum is (4686). In exams, add terms carefully when the ratio is large.

In option A, each term is obtained by multiplying the preceding term by \(-\frac{1}{2}\): \(-8/16=4/-8=-2/4\). Hence it is a GP with a negative common ratio. In option B, the ratios are not constant. Exam tip: check at least two consecutive ratios.

The first term is (216) and the ratio is \(\frac{1}{3}\), so the correct rule is \(216\cdot\left(\frac{1}{3}\right)^{n-1}\). In exams, write the fractional ratio in a decreasing GP.

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 \(n-1\).