Geometric Progression

Given \(a_n=6\cdot2^{n-1}\), for \(n=1\), \(a_1=6\cdot2^0=6\); for \(n=2\), \(a_2=6\cdot2^1=12\); and for \(n=3\), \(a_3=6\cdot2^2=24\). Therefore, the first three terms are \((6, 12, 24)\). Option B misses the initial coefficient 6. Exam tip: always substitute \(n=1\) first to check the first term.

The first term is (7) and the ratio is (4) so (a_n=7\cdot4^{n-1}). In exams match both the first term and the ratio.

The common ratio is (5) so the missing term is (100\times5=500). In exams continue the same ratio.

The ratio of consecutive terms is (24\div72=\frac{1}{3}). In exams write the fractional ratio for a decreasing geometric progression.

After the fourth term (128) the fifth is (512) and the sixth is (2048). In exams include the first term while counting.

A sequence is a geometric progression when the quotient of every term and its preceding term is constant. Here, 42 ÷ 14 = 3, 126 ÷ 42 = 3, and 378 ÷ 126 = 3. Since the same ratio occurs throughout the displayed sequence, it is a geometric progression with common ratio 3. Therefore option A is correct. Option B mistakes the first term for the common ratio. Option C uses an irrelevant condition: the terms do not need to be equal; they need to have a constant multiplicative ratio. Option D is false because the terms increase, rather than decrease, as each term is tripled.

In a geometric progression, the first term is a, and each successive term is found by multiplying the preceding term by the common ratio r. Here, a = 4 and r = 7: 4, 4×7 = 28, 28×7 = 196, and 196×7 = 1372. Therefore, (4, 28, 196, 1372) is correct. Option A incorrectly treats 7 as the second term; it is the common ratio. Exam tip: write a first, then multiply each term by r.

The first term is (100) and the ratio is \(\frac{1}{5}\) so \(a_n=100\cdot\left(\frac{1}{5}\right)^{n-1}\). In exams use the fractional ratio in a decreasing sequence.

In a geometric progression, the first term is \(a\), so \(a=3\). The common ratio \(r\) is found by dividing a term by the preceding term: \(r=\frac{18}{3}=6\). Checking, \(18\times6=108\) and \(108\times6=648\). Hence, \(a=3,\ r=6\). In option D, the ratio is correct, but 18 is not the first term. Exam tip: identify the first term as \(a\), then divide consecutive terms to find \(r\).

In a geometric progression, each term is obtained by multiplying the previous term by a fixed number, so consecutive-term ratios are equal. For 2, 6, 18, 6/2 = 18/6 = 3. Equal differences identify an arithmetic progression. In exams, check ratios first.

This is a geometric sequence in which each term is 3 times the preceding term: 6, 18, 54, 162, 486. Therefore, 486 is the fifth term. The sixth term would be 1458, so the sixth-term option is incorrect. Exam tip: Write the terms in order and count their positions starting from 1.

The common ratio in a geometric progression is found by dividing a term by the preceding term. Since a₁ = 12 and a₂ = 48, calculate r = a₂ ÷ a₁ = 48 ÷ 12 = 4. Therefore option C is correct. The result can be checked by multiplying the first term by the ratio: 12 × 4 = 48, exactly matching the second term. A ratio of 2 would produce 24, a ratio of 3 would produce 36, and a ratio of 6 would produce 72. Thus none of the other choices connects the two given terms. The order of division matters: later term divided by earlier term, not the reverse.

The common ratio is (2) so the next term is (104\times2=208). In exams multiply the last known term by the ratio.

The common ratio is (\frac{1}{2}) so the fifth term is (30\times\frac{1}{2}=15). In exams apply fractional ratios carefully.

The common ratio is the quotient of a term and the immediately preceding term. For the algebraic sequence z, 5z, 25z, divide the second term by the first: r = 5z ÷ z = 5, assuming z is nonzero so that the quotient is defined. The next pair confirms the same value: 25z ÷ 5z = 5. Therefore option B is correct. The variable z cancels because it is a common non-zero factor in both terms. Option A is the first term, not the multiplier. Option C is the coefficient of the third displayed term, while 125 is an unrelated value and does not describe the step from one term to the next. The constant multiplier, not the term itself, is required.

Consecutive terms are equal so the ratio is (11\div11=1). In exams a non-zero constant sequence can also be a geometric progression.

When the common ratio is (1) all terms remain (17). In exams (r=1) means a constant geometric progression.

The ratio of (100) and (500) is (5) so the missing term is (100\div5=20). In exams apply the common ratio backward too.

In the formula (a_n=9\cdot3^{n-1}) the common ratio is (3). In exams identify (r) in (ar^{n-1}).

In a GP, every term is a fixed multiple of the preceding term; this fixed number is the common ratio. Compare two successive ratios to check it. Equal differences describe an arithmetic progression, not a GP.