Geometric Progression

A sequence is a geometric progression when the ratio of every term to the term immediately before it is constant. The terms do not need to be equal; instead, they are repeatedly multiplied by the same number. This constant number is called the common ratio.

For the given sequence, \(20\div10=2\), \(40\div20=2\), and \(80\div40=2\). Since all consecutive ratios are equal to 2, the sequence is a geometric progression with common ratio 2. Therefore, option A is correct. The value 10 is the first term, not the ratio, and the sequence is increasing rather than decreasing.

In a geometric progression, each term is obtained by multiplying the preceding term by the common ratio r. Here, the first term is 2 and r = 5: 2, 2 × 5 = 10, 10 × 5 = 50, and 50 × 5 = 250. Therefore, the correct sequence is (2, 10, 50, 250). Option C incorrectly begins with 5 instead of the given first term. Exam tip: write the first term first, then multiply successively by r.

The first term is (27) and the ratio is \(\frac{1}{3}\), so \(a_n=27\cdot\left(\frac{1}{3}\right)^{n-1}\). In exams, use the fractional ratio in a decreasing sequence.

The governing concept is the explicit formula for the general term of a geometric progression: a_n = a_1r^(n-1). To find the fourth term, substitute n = 4 into the given rule: a_4 = 16(1/2)^(4-1) = 16(1/2)^3. Since (1/2)^3 = 1/8, a_4 = 16 × 1/8 = 2. Therefore option B is correct. The sequence generated by the rule begins 16, 8, 4, 2, which provides a quick independent check. Option C would result from applying the halving operation only twice, while option D corresponds to the second term. Option A applies one extra halving. The exponent n - 1 correctly counts the number of ratio applications needed to reach the nth term.

In a geometric progression, \(a\) is the first term, so \(a=1\). The common ratio \(r\) is obtained by dividing a term by the preceding term: \(r=\frac{5}{1}=5\). Therefore, \(a=1,\ r=5\) is correct. \(r=25\) is incorrect because it compares the first and third terms, not consecutive terms. Exam tip: find \(r\) by dividing the second term by the first term.

The \(n\)th term of a geometric progression is \(a_n=ar^{n-1}\). Therefore, \(a_5=7\times2^{5-1}=7\times16=112\). Hence, 112 is correct. The value 56 may result from incorrectly using \(2^3\). Exam tip: in the \(n\)th-term formula, the exponent of \(r\) is always \(n-1\).

This is a geometric sequence in which each term is twice the previous term: 4, 8, 16, 32, 64. Therefore, 64 is the fifth term. The fourth term is 32, so option A is not correct. Exam tip: while finding a term’s position, count the first given term as the first term.

In option A, each term is multiplied by 3: 6/2 = 3 and 18/6 = 3. Hence, the common ratio is constant, so it is a geometric progression. In exams, check whether ratios of consecutive terms are equal.

The common ratio is (2), so the next term is (72\times2=144). In exams, multiply the last known term by the ratio.

In a geometric progression, the ratio of consecutive terms is constant; each term is obtained by multiplying the previous term by the same fixed number, called the common ratio. Adding a fixed number describes an arithmetic progression. Exam tip: check ratios, not differences.

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 direct answer is A: \(a_n=20\cdot(1/2)^{n-1}\). A geometric progression has a constant ratio between consecutive terms. Here 10/20=1/2, 5/10=1/2 and (5/2)/5=1/2, so the first term is a=20 and the common ratio is r=1/2. The general formula is \(a_n=ar^{n-1}\), hence \(a_n=20(1/2)^{n-1}\). Checking it: n=1 gives 20, n=2 gives 10, n=3 gives 5 and n=4 gives 5/2. Option A matches every term. Option B uses ratio 2 and would produce 20, 40, 80, not the given decreasing sequence. Option C is a subtraction pattern, not a geometric formula, and does not give the terms. Option D gives 10, 20, 30, so it also fails. Memory cue: in a GP identify the first term and multiply repeatedly by the same ratio.

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.