Geometric Progression

Given \(a_n=5\cdot2^{n-1}\) and \(a_n=320\), we get \(5\cdot2^{n-1}=320\). Thus, \(2^{n-1}=64=2^6\). Equating exponents gives \(n-1=6\), so \(n=7\). If \(n=6\), the term would be \(5\cdot2^5=160\), not 320. Exam tip: first divide by the constant factor to isolate the exponential part.

The direct answer is Option B: r=
(1/3) and the progression decreases. A geometric progression has one fixed multiplier between consecutive terms. Divide the second term by the first: 6/18=
(1/3). Check the next steps: 6 times
(1/3)=2, and 2 times
(1/3)=
(2/3). Thus every term is one third of the preceding term. Since the terms are positive and the multiplier lies between 0 and 1, each new term is smaller, so the progression decreases. Option A says r=3 and increasing. It is wrong because the actual ratio is
(1/3), not 3; also the terms do not increase. Option B is correct on both parts. Option C says r=-3 and alternating signs. It is wrong because the ratio is positive and all displayed terms have the same positive sign. Option D says it is not a geometric progression, but the same ratio occurs at every step, so it clearly is one. Exam cue: calculate “next term divided by previous term”; do not reverse the division. A positive ratio below 1 usually gives a decreasing positive progression.

For a geometric progression, \(a_n=a_m r^{n-m}\). Hence, \(a_5=a_2r^{5-2}=24\left(\frac{1}{2}\right)^3=24\times\frac{1}{8}=3\). Note that \(a_4=6\), so 6 is a close distractor; to reach \(a_5\) from \(a_2\), multiply by the common ratio three times.

The terms are (2,10,50,250,1250,6250), so the greatest term below (6000) is (1250). The next term crosses the limit.

In a GP, dividing each term by its preceding term gives the same common ratio: \(a_{n+1}/a_n=r\). Option A describes an arithmetic progression with a constant difference. Exam tip: compare ratios of consecutive terms, not differences.

The general term of a GP is \(a_n=a_1r^{n-1}\). Thus, \(a_4=9r^3=72\), so \(r^3=8\) and the positive value of \(r\) is 2. Option 8 is the value of \(r^3\), not of \(r\). Exam tip: in \(a_n\), the exponent of \(r\) is always \(n-1\).

(\frac{a_5}{a_3}=r^2=3^2=9). In the same progression, the ratio of terms depends on the difference of positions.

In a geometric progression, the ratio of consecutive terms is constant. Here, \(\frac{48}{12}=4\). Therefore, \(\frac{12}{x}=4\), which gives \(x=3\). If \(x=4\), the two ratios would be \(3\) and \(4\), so the terms would not form a GP. Exam tip: for three consecutive GP terms \(a,b,c\), you may also use \(b^2=ac\).

The first (5) terms are (12,6,3,\frac{3}{2},\frac{3}{4}), and their sum is (\frac{93}{4}). Add fractional terms using a common denominator.

In a geometric progression, \(a_n=a_1r^{n-1}\). Therefore, \(a_3=11\times3^2=99\) and \(a_5=11\times3^4=891\). Hence, \(a_3+a_5=99+891=990\). Options such as 1000 do not follow from calculating the terms with the required powers of the common ratio. Exam tip: for the \(n\)th term, use the exponent \(n-1\).

In a geometric progression, the \(n\)th term is \(a_n=a_1r^{n-1}\). Therefore, \(a_4=8\left(\frac{3}{2}\right)^{4-1}=8\times\frac{27}{8}=27\). Option 36 may result from using an incorrect power of the common ratio or multiplying incorrectly. Exam tip: in \(a_n\), the exponent of \(r\) is always \(n-1\), not \(n\).

(a_1=\frac{14}{5}) and (a_3=70), so the sum is (\frac{364}{5}). Use the ratio both backward and forward.

(x^2=16\cdot81=1296), so positive (x=36). The square of the middle term equals the product of the outer terms.

In a geometric progression, each term is obtained by multiplying the preceding term by the same constant ratio. In option B, \(18/6=3\) and \(54/18=3\), so the common ratio is 3 and the terms form a GP. In option A, the ratios are 2 and \(5/2\), so it is not a GP. Exam tip: for three terms \(a,b,c\), check whether \(b/a=c/b\).

Direct answer: Option D, 3125. In a GP, the nth term is obtained by multiplying the first term by r repeatedly: a_n=a_1r^{n-1}. Since a_1=5 and a_4=625, we have 625=5r^3. Dividing by 5 gives r^3=125, so the positive ratio is r=5. The next term is a_5=a_4r=625×5=3125. Option A, 1250, would use a factor of 2 and does not follow from the given ratio. Option B, 1875, would use a factor of 3. Option C, 2500, would use a factor of 4. Only option D uses the required factor 5. The condition that r is positive removes any negative cube-root concern. Memory cue: find the ratio from two terms, then multiply the last known term once.

In option A, the successive ratios are \(6/(-3)=-2\), \((-12)/6=-2\), and \(24/(-12)=-2\), so it is a GP. In option C, the ratios change. Exam tip: compare ratios of consecutive terms.

Dividing \(a_n^2=a_{n-1}a_{n+1}\) by \(a_{n-1}a_n\) gives \(\frac{a_n}{a_{n-1}}=\frac{a_{n+1}}{a_n}\). Thus consecutive terms have the same ratio, so the sequence is a GP. Option A describes an AP with a constant difference. Exam tip: test the ratio, not the difference, for a GP.

For three consecutive terms of a geometric progression, the square of the middle term equals the product of the first and third terms. Thus, \(x^2=6\times150=900\). Hence \(x=\pm30\), but the progression is stated to be positive, so \(x=30\). A value such as \(36\) does not give equal common ratios. Exam tip: for consecutive GP terms \(a,b,c\), use \(b^2=ac\) directly.

(a_5=a_1r^4), so (1792=7r^4) and (r^4=256), hence (r=4). When positive ratio is asked, take the positive root.

For consecutive GP terms, the common-ratio condition gives \(q/p=r/q\). Cross-multiplying gives \(q^2=pr\). The relation \(p+r=2q\) belongs to an AP. Exam tip: square the middle term and compare it with the product of the outer terms.

In (5,15,45,135,405,1215,\ldots), (a_3=45) and (a_6=1215). Check each option up to the required positions.

Given \(a_n=4\cdot3^{n-1}\), \(a_4=4\cdot3^{3}=108\) and \(a_6=4\cdot3^{5}=972\). Therefore, \(a_4+a_6=108+972=1080\). The option 972 is only the value of \(a_6\), not the required sum. Exam tip: while finding \(a_n\), remember that the exponent is \(n-1\).

The terms are (384,192,96,48,24,12,6), so (6) is the seventh term. In a decreasing progression, multiply repeatedly by the ratio.

(a_7=a_3r^4), so (576=36r^4) and (r=2), hence (a_1=\frac{36}{2^2}=9). Divide by the ratio to move backward.

For the middle term, ((3x)^2=(x+2)(6x+4)), which gives (x=4). In three GP terms, the square of the middle term equals the product of the extremes.