Geometric Progression

The terms are (9,18,36,72,144,288,576,1152), so the greatest term below (600) is (576). In boundary questions, check the next term too.

In (32,16,8,4,2,\ldots), (r=\frac{1}{2}) and the fifth term is (2). Check both first and fifth terms in the options.

For (r=3), the terms are (4,12,36,108,324) and the sum is (484). For small options, forming the terms is quick.

The first term is \(a=5\), and the common ratio is \(r=\frac{-20}{5}=-4\). Thus, \(a_3=5(-4)^2=80\) and \(a_5=5(-4)^4=1280\). Therefore, \(a_5+a_3=1280+80=1360\). An answer such as \(1120\) can result from using the wrong power or term position with the negative common ratio. Exam tip: use \(a_n=ar^{n-1}\) and check the exponent before calculating.

Given \(a_n=96\left(\frac{1}{2}\right)^{n-1}\), \(a_3=96\left(\frac{1}{2}\right)^2=24\) and \(a_6=96\left(\frac{1}{2}\right)^5=3\). Therefore, \(a_3+a_6=24+3=27\). The option 30 may result from using an incorrect exponent instead of \(n-1\). Exam tip: write the exponent \(n-1\) first before evaluating each term.

The nth term of a geometric progression is \(a_n=a_1r^{n-1}\). Therefore, \(a_4=\frac{7}{2}\times4^{4-1}=\frac{7}{2}\times64=224\). The value \(448\) usually results from an error in evaluating the power of \(4\) or in multiplication. Exam tip: for the fourth term, the exponent of the common ratio is always \(4-1=3\).

The terms are (3,15,75,375,1875,9375,46875), so the greatest term below (10000) is (9375). In boundary questions, check the next term too.

In a geometric progression, \(a_n=a_1r^{n-1}\). Thus, \(a_4=9\times3^3=243\) and \(a_5=9\times3^4=729\). Therefore, \(a_4+a_5=243+729=972\). A value such as \(1215\) can result from using an incorrect power or term value. Exam tip: for the \(n\)th term, use the exponent \(n-1\).

The first term is 12 and the common ratio is 3. Thus, \(a_4=12\times 3^3=324\) and \(a_6=12\times 3^5=2916\). Therefore, \(a_6-a_4=2916-324=2592\). A value such as 2484 can result from an error in subtraction. Exam tip: while using \(a_n=ar^{n-1}\) for a GP, remember that the exponent is \(n-1\).

Given \(a_n=125\left(\frac{1}{5}\right)^{n-1}\), \(a_2=125\left(\frac{1}{5}\right)^1=25\) and \(a_4=125\left(\frac{1}{5}\right)^3=1\). Hence, \(a_2+a_4=25+1=26\). The option 30 is incorrect because \(a_4\) is 1, not 5. Exam tip: for a particular term, substitute the value of \(n\) first and then evaluate the exponent carefully.

In a geometric progression, a gap of four indices gives \,a_8=a_4r^{8-4}=a_4r^4\,. Thus \,4374=54r^4\, and \,r^4=4374/54=81\,. Hence \,r=3\, because r is stated to be positive. Although 9 may seem tempting, it is not r; it is related to the value of \,r^4\,. Exam tip: use the difference between the term indices to determine the exponent of r.

In (64,-16,4,-1,\ldots), each term is multiplied by (-\frac{1}{4}). Divide the next term by the previous term to find the ratio.

In a GP, \(a_4=a_2r^2\). Thus, \(250=10r^2\), so \(r^2=25\). Since \(r\) is positive, \(r=5\). Now \(a_1=a_2/r=10/5=2\), and \(a_5=a_4r=250\times5=1250\). Therefore, \(a_1+a_5=2+1250=1252\). Although \(r=-5\) also gives \(r^2=25\), the question specifically requires positive \(r\). Exam tip: use the difference between term indices to determine the required power of \(r\).

Given \(8\cdot3^{n-1}=1944\). Dividing both sides by 8 gives \(3^{n-1}=243\). Since \(243=3^5\), we get \(n-1=5\), hence \(n=6\). If 5 were chosen, the exponent would be \(n-1=4\), so the term would not be 1944. Exam tip: first remove the coefficient, then compare exponents with the same base.

The direct answer is Option B: r=
(1/2) and the progression decreases. Find the common ratio by dividing a term by the term immediately before it: 20/40=
(1/2), 10/20=
(1/2), and 5/10=
(1/2). The same ratio appears each time, so the sequence is geometric. Because the ratio is positive and less than 1, each positive term becomes half as large: 40, 20, 10, 5 and then 2.5. Thus the progression decreases. Option A says r=2 and increasing; it reverses the division and also contradicts the visible decrease. Option B gives both the correct ratio and the correct direction. Option C says r=-2 and alternating signs; that would produce signs such as positive, negative, positive, but all displayed terms are positive. Option D says it is not a geometric progression, but the constant ratio proves that it is. Exam cue: use “new term divided by old term”; a positive ratio between 0 and 1 means decreasing positive terms.

The direct answer is option A: \(a_6=\frac{5}{3}\). In a geometric progression, moving from one term to a later term means multiplying by the common ratio once for each step. From the third term to the sixth term there are three steps, so \(a_6=a_3r^3\). Substitute the given values: \(a_6=45\left(\frac13\right)^3=45\times\frac1{27}=\frac{45}{27}=\frac53\). Therefore option A is correct. Option B, 5, would result from an incomplete or incorrect division and is too large for three multiplications by \(\frac13\). Option C, 15, is what one might get after dividing only once, so it ignores two of the three steps. Option D, 135, increases the value and would be inconsistent with a ratio smaller than 1. The terms move downward: \(a_4=15\), \(a_5=5\), and \(a_6=\frac53\), which confirms the result. Memory cue: between term numbers \(m\) and \(n\), the power is \(n-m\), not \(n\) itself.

The terms are (11,22,44,88,176,352,704,1408), so the greatest term below (800) is (704). The next term crosses the limit.

For a geometric progression, the relation between two terms is determined by the common ratio. From the second term to the sixth term there are four equal steps, so \\(a_6=a_2r^4\\). Substituting the given values gives \\(1215=15r^4\\), and therefore \\(r^4=81\\). Since the ratio is stated to be positive, the suitable fourth root is \\(r=3\\), not \\(r=-3\\).

The seventh term is one more multiplication by the ratio after the sixth term. Thus \\(a_7=a_6r=1215\\times3=3645\\). Therefore option B is correct. The condition that the ratio is positive is important because the equation for \\(r^4\\) alone would also allow a negative fourth root, but that possibility is excluded by the question.

The first four terms are (3,12,48,192), and their product is (589824). Multiply each term carefully in product questions.

The general-term formula of a GP is \(a_n=a_1r^{n-1}\). Thus, \(a_4=10r^3=270\), so \(r^3=27\) and the positive value is \(r=3\). If 2 were used, the fourth term would be \(10\times2^3=80\), not 270. Exam tip: in \(a_n\), the exponent of \(r\) is always \(n-1\).

(\frac{a_7}{a_4}=r^3=4^3=64). In the same progression, the ratio of terms depends on the difference of positions.

In a geometric progression, the ratios of consecutive terms are equal. Here, the common ratio is \(\frac{96}{24}=4\), so \(\frac{24}{x}=4\). Hence, \(x=\frac{24}{4}=6\). Equivalently, for three consecutive GP terms, the square of the middle term equals the product of the outer terms: \(24^2=96x\), giving \(x=6\). If \(x=8\), the ratios are \(24/8=3\) and \(96/24=4\), which are not equal. Exam tip: For consecutive GP terms \(a,b,c\), use \(b^2=ac\) directly.

The first (5) terms are (36,18,9,\frac{9}{2},\frac{9}{4}), and their sum is (\frac{279}{4}). Add fractional terms using a common denominator.

For a geometric progression, the nth term is
. Hence,
and
. Therefore,
. The nearby option 660 does not result when the correct powers of the common ratio are used. Exam tip: for the nth term, the exponent of
is always
, not
.

In a geometric progression, the \(n\)th term is \(a_n=a_1r^{n-1}\). Therefore, \(a_4=27\left(\frac{2}{3}\right)^{4-1}=27\left(\frac{2}{3}\right)^3=27\times\frac{8}{27}=8\). The value 6 can result from incorrectly multiplying by the common ratio only twice. Exam tip: in \(a_n\), the exponent of the common ratio is always \(n-1\).