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SubjectsMathematics
गुणोत्तर श्रेणी
In this Class 9 Mathematics topic from Sequences and Progressions, students explore sequences in which each term is obtained by multiplying the preceding term by a fixed number. They learn to identify the common ratio, distinguish a geometric progression from other patterns, write its terms, and use the general term to find a specific position in the sequence. Examples help connect the idea with repeated growth, decrease, and everyday numerical patterns.
TOPIC PRACTICE
Up to 24 questions from this page. Select your focus, then start.
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.
The nth term of a geometric progression is \(a_n=a_1r^{n-1}\). Therefore, \(a_3=7\times5^{3-1}=7\times25=175\). The value 35 is only the second term, since \(a_2=7\times5\). Exam tip: for the third term, use \(r^2\) in the general-term formula.
This is of the form (3\cdot2^{n-1}) so the tenth term is (3\cdot2^9=1536). In exams identify powers to solve quickly.
Given \(a_n=2\cdot5^{n-1}\), substitute \(n=4\): \(a_4=2\cdot5^{4-1}=2\cdot5^3=2\cdot125=250\). Hence, option C is correct. Option D can result from an error in multiplying after evaluating \(5^3\). Exam tip: substitute the term number first, and then evaluate the exponent carefully.
The direct answer is A: \(a_n=60\cdot(1/2)^{n-1}\). The defining feature of a geometric progression is a fixed multiplier. Calculate the consecutive ratios: 30/60=1/2, 15/30=1/2 and (15/2)/15=1/2. Therefore the first term is 60 and the common ratio is 1/2. Applying \(a_n=ar^{n-1}\) gives option A. It checks correctly: n=1 gives 60, n=2 gives 30, n=3 gives 15 and n=4 gives 15/2. Option B uses ratio 2 and would produce increasing terms 60, 120, 240. Option C subtracts the index and gives 59, 58, 57, not the sequence. Option D gives 30, 60, 90, so it is also inconsistent. A is correct because it reproduces every listed term. Memory trick: for repeated halving, write the ratio as \(1/2\) and use exponent n-1.
The square of the middle term is (10\times90=900) so the positive (x=30). 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 (486\times3=1458). In exams multiply the last known term by the ratio.
A geometric progression with first term 8 and common ratio 3 must begin with 8, and each later term must be three times the preceding term. Option A satisfies both requirements: 24 = 8 × 3, 72 = 24 × 3, and 216 = 72 × 3. Therefore option A is correct. Option B has ratio 3 but begins with 3, so its first term is wrong. Option C begins with 8 but adds 3 each time, making it an arithmetic progression with common difference 3. Option D follows ratio 3 but begins with 24, which would be the second term of the required progression. Both the starting value and repeated multiplication must be checked.
The nth term of a geometric progression is \(T_n=ar^{n-1}\). Therefore, \(T_6=12\times2^{6-1}=12\times2^5=12\times32=384\). The value \(288\) can result from an error in evaluating the power or multiplication. Exam tip: for the sixth term, always use the exponent \(6-1=5\).
The third term is (a_3=a_2\times r=15\times3=45). In exams multiply the previous term by the ratio to get the next term.
The governing concept is the definition of a geometric progression: every term must be obtained by multiplying the preceding term by the same common ratio. We must check two independent conditions, not just one. In option A, the sequence is 2, 10, 50, 250, so its third term is 50. Also, 10/2 = 5, 50/10 = 5, and 250/50 = 5; therefore its common ratio is 5. Option B has ratio 5 but its third term is 125. Option C also has ratio 5, but its third term is 250. Option D has ratio 5, but its third term is 25. Hence only option A satisfies both requirements, so A is unambiguously correct.
A geometric progression has the same ratio between every pair of consecutive terms, and a_4 means the fourth listed term. In option A, the sequence is 12, 24, 48, 96, so each term is twice the preceding term and r = 2. Its fourth term is exactly 96, satisfying both conditions. Option B has ratio 2 but its fourth term is 192. Option C also has ratio 2, but its fourth term is 48. Option D has ratio 2 as well, yet its fourth term is 768. Therefore only option A is correct. The question requires checking both the common ratio and the position of the specified term, not just one of them.
QUIZ COMPLETE
