Geometric Progression

The common ratio is (\frac{1}{2}) so the fifth term is (12\times\frac{1}{2}=6). In exams simplify fractions.

The terms are (2,6,18,54,162) so (162) is the fifth term. In exams list small terms in order to check.

For \(n=1\), \(a_1=4\cdot3^0=4\); for \(n=2\), \(a_2=4\cdot3^1=12\); and for \(n=3\), \(a_3=4\cdot3^2=36\). Therefore, the first three terms are \((4, 12, 36)\). In \((3, 9, 27)\), the first term is 3, but the given formula gives \(a_1=4\). Exam tip: substitute \(n=1\) first to verify the initial term.

The first term is (9) and the ratio is (4) so (a_n=9\cdot4^{n-1}). In exams match both the first term and the ratio.

The governing property is a constant common ratio between consecutive terms. From the first two terms, r = 15 ÷ 3 = 5; this is confirmed because 75 ÷ 15 = 5. To obtain the missing fourth term, multiply the third term by the same ratio: 75 × 5 = 375. Thus option C is correct. Option A would multiply 75 by 3, not by the established ratio; option B does not preserve the ratio; and option D would require a ratio of 6. Continuing a geometric progression means applying the same multiplier at every step, not adding a fixed number or choosing a convenient larger value.

The ratio of consecutive terms is (24\div48=\frac{1}{2}). In exams write the fractional ratio for a decreasing geometric progression.

After the fourth term (320) the fifth is (1280) and the sixth is (5120). In exams include the first term while counting.

The defining feature of a geometric progression is a constant multiplier between consecutive terms. To test the sequence, divide each term by the preceding term. If the same quotient is obtained each time, the sequence is geometric, and that quotient is its common ratio.

For this sequence, \(36\div12=3\), \(108\div36=3\), and \(324\div108=3\). Thus every term is obtained by multiplying the previous term by 3. The sequence is therefore a geometric progression with common ratio 3, so option A is correct. The number 12 is the first term, while the sequence is increasing, not decreasing.

In a geometric progression, each term is obtained by multiplying the preceding term by the common ratio. Here, the first term is 3 and r = 6: 3, 3×6 = 18, 18×6 = 108, and 108×6 = 648. Therefore, (3, 18, 108, 648) is correct. In option A, the terms are multiplied by 2, not by 6. Exam tip: Write the first term first, then multiply each successive term by r.

The first term is (64) and the ratio is \(\frac{1}{2}\) so \(a_n=64\cdot\left(\frac{1}{2}\right)^{n-1}\). In exams use the fractional ratio in a decreasing sequence.

In a geometric progression, \(a\) is the first term, so \(a=2\). The common ratio \(r\) is the quotient of consecutive terms: \(r=\frac{10}{2}=5\). This is confirmed by \(50\div10=5\) and \(250\div50=5\). In option D, the ratio is correct, but the first term is 2, not 10. Exam tip: identify \(a\) from the first term and find \(r\) by dividing the second term by the first.

In a geometric progression, each term is obtained by multiplying the previous term by the same constant ratio. In option B, 6/3 = 12/6 = 24/12 = 2. Options A and D are arithmetic progressions because their common difference is constant. Exam tip: divide consecutive terms to check the common ratio.

The terms in order are 3, 6, 12, 24, and 48. Therefore, 48 is the fifth term. The sixth term would be 96, so it is not correct. Exam tip: always count the given first term as term 1.

In a geometric progression, the common ratio is the ratio of consecutive terms. Thus, \(r=\frac{a_2}{a_1}=\frac{32}{8}=4\). Therefore, 4 is correct. The number 8 is the first term, not the common ratio. Exam tip: divide the second term by the first term to find the common ratio.

The common ratio is (2) so the next term is (88\times2=176). In exams multiply the last known term by the ratio.

The common ratio is (\frac{1}{2}) so the fifth term is (20\times\frac{1}{2}=10). In exams apply fractional ratios carefully.

Direct answer: the common ratio is 4, so option C is correct. In a geometric progression, divide any nonzero term by the term immediately before it. Using the first two terms, r = 4y/y = 4, provided y is not zero so that the ratio is defined. The next pair confirms this: 16y/(4y) = 4. Therefore every term is obtained by multiplying the previous term by 4: y × 4 = 4y and 4y × 4 = 16y. Option A would mean the terms remain unchanged, which they do not. Option B would give the second term as 2y, not 4y. Option C gives both displayed transitions correctly. Option D is the third coefficient, not the ratio between consecutive terms. A common confusion is to choose a number appearing in a term rather than calculate a quotient; always compare consecutive terms by division.

Consecutive terms are equal so the ratio is (8\div8=1). In exams a non-zero constant sequence can also be a geometric progression.

When the common ratio is (1) all terms remain (13). In exams (r=1) means a constant geometric progression.

The ratio of (50) and (250) is (5) so the missing term is (50\div5=10). In exams apply the common ratio backward too.