Geometric Progression

The common ratio is (\frac{1}{2}), so the fifth term is (25\times\frac{1}{2}=\frac{25}{2}). In exams, apply fractional ratios carefully.

The governing concept is the definition of a geometric progression: the quotient of every term and its preceding term must be the same constant, called the common ratio. Using the first two terms, r = 3x ÷ x = 3, provided x is not zero. The next pair confirms the result because 9x ÷ 3x = 3 as well. Therefore, each term is obtained by multiplying the previous term by 3, and option C is correct. Option A is merely the variable appearing in the first term, not the multiplying factor. Option B is not the quotient of consecutive terms, while option D is the coefficient in the third term rather than the common ratio. The pattern is x, 3x, 9x, 27x, and so on.

Consecutive terms are equal, so the ratio is (5\div5=1). In exams, a non-zero constant sequence can also be a geometric progression.

When the common ratio is (1), all terms remain (11). In exams, (r=1) means a constant geometric progression.

The ratio of (27) and (81) is (3), so the missing term is (27\div3=9). In exams, apply the common ratio backward too.

In the formula (a_n=4\cdot2^{n-1}), (2) is the common ratio. In exams, identify (r) in (ar^{n-1}).

The common ratio of this geometric progression is \(r=36/12=3\). Since the fourth term is 324, the fifth term is \(a_5=324\times 3=972\). The value 1296 would come from multiplying 324 by 4, which is not the common ratio. Exam tip: obtain each next term by multiplying the previous term by the common ratio.

In a GP, each next term is obtained by multiplying by the same common ratio. A constant difference identifies an AP, not a GP. Exam tip: compare ratios of consecutive terms.

In (64,16,4,1), each term is multiplied by (\frac{1}{4}). In exams, write the ratio of a decreasing geometric progression as a fraction.

This is of the form (2^n), so the tenth term is (2^{10}=1024). In exams, identify powers to solve quickly.

The general term is \(a_n=3^n\). Substituting \(n=4\), we get \(a_4=3^4=3\times3\times3\times3=81\). Therefore, 81 is correct. Note that \(3^3=27\), so choosing 27 would use exponent 3 instead of 4. Exam tip: substitute the term number carefully before evaluating the power.

The first term is (20) and the ratio is \(\frac{1}{2}\). In exams, put the correct fractional ratio in \(ar^{n-1}\).

The square of the middle term is (6\times54=324), so the positive (x=18). In exams, for three GP terms, the square of the middle term equals the product of the outer terms.

The common ratio is (3), so the fifth term is (405\times3=1215). In exams, multiply the last known term by the ratio.

A geometric progression with first term 4 and common ratio 3 must start at 4, and every next term must be formed by multiplying the previous term by 3. Option A gives 4, then 4 × 3 = 12, 12 × 3 = 36, and 36 × 3 = 108. Thus it satisfies both required conditions. Option B has ratio 2 and begins with 3, so it fails both tests. Option C begins with 4 but increases by adding 4; its ratios are not constant at 3. Option D has successive ratio 3, but its first term is 12 rather than 4. Therefore, only option A is a geometric progression with the specified first term and common ratio. The check must consider both conditions, not just the ratio.

The governing concept is that consecutive terms of a geometric progression have a constant quotient. Divide the second term by the first: \(r=\frac{6}{3}=2\). The remaining terms confirm this, because \(\frac{12}{6}=2\) and \(\frac{24}{12}=2\). Thus option A is correct. A ratio of 3 would produce 3, 9, 27, and so on, while ratios 4 and 6 would produce much larger successive terms. Although the terms visibly double, the formal method is to divide one term by the preceding term and verify that the quotient remains constant.

The \(n\)th term of a GP is \(a_n=ar^{n-1}\). Here, \(a=5\), \(r=3\), and \(n=4\), so \(a_4=5\times3^{4-1}=5\times27=135\). Therefore, 135 is correct. A value such as 90 results from using the exponent or term position incorrectly. Exam tip: to reach the fourth term, multiply the common ratio \(r\) three times.

In (4,12,36,108,\ldots), each term is multiplied by (3). In a geometric progression consecutive ratios are equal.

The common ratio is (\frac{27}{81}=\frac{1}{3}). In a decreasing geometric progression the ratio can be less than (1).

Given \(a_n=2\cdot4^{n-1}\), put \(n=5\): \(a_5=2\cdot4^{5-1}=2\cdot4^4=2\cdot256=512\). Therefore, 512 is correct. The value 256 is only \(4^4\); the initial factor 2 must also be multiplied. Exam tip: substitute the given value of \(n\) into \(n-1\) first.