Geometric Progression

The ratios (4\div1=4) and (9\div4=\frac{9}{4}) are not equal. In exams, equal ratio is necessary for a geometric progression.

In a geometric progression, the common ratio \(r\) is found by dividing any nonzero term by the term immediately before it. Using the first two terms gives \(r=\frac{12}{4}=3\). This is confirmed by the next pairs: \(\frac{36}{12}=3\) and \(\frac{108}{36}=3\). Therefore, option B is correct. Option A would describe doubling, option C would require each term to be four times the previous one, and option D would require a factor of six. Since every displayed transition is multiplication by 3, the sequence is consistent with \(r=3\).

The \(n\)th term of a geometric progression is \(T_n=ar^{n-1}\). Hence, \(T_4=6\times2^{4-1}=6\times2^3=48\). The value 24 is the third term, since \(T_3=6\times2^2=24\). Exam tip: the exponent of the common ratio in the \(n\)th term is always \(n-1\).

The common ratio is (\frac{1}{2}), so the fifth term is (\frac{25}{2}\times\frac{1}{2}=\frac{25}{4}). In exams, simplify fractions.

The terms are (1,2,4,8,16), so (16) is the fifth term. In exams, list small terms in order to check.

Given \(a_n=3\cdot2^{n-1}\): for \(n=1\), \(a_1=3\cdot2^0=3\); for \(n=2\), \(a_2=3\cdot2^1=6\); and for \(n=3\), \(a_3=3\cdot2^2=12\). Hence, the first three terms are \((3, 6, 12)\). The sequence \((3, 9, 27)\) has common ratio 3, whereas this sequence has common ratio 2. Exam tip: substitute \(n=1\) first and check the exponent \(n-1\) carefully.

The first term is (8) and the ratio is (3), so (a_n=8\cdot3^{n-1}). In exams, match both the first term and the ratio.

The common ratio is (5), so the missing term is (50\times5=250). In exams, continue the same ratio.

The ratio of consecutive terms is (16\div32=\frac{1}{2}). In exams, write the fractional ratio for a decreasing geometric progression.

The fifth term is (486), not the sixth; the asked term should be counted carefully. In exams, include the first term while counting.

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.