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T12 of the series –128, 64, –32, ….is (a) – 1/16 (b) 16 (c) 1/16 (d) none of these?
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T12 of the series –128, 64, –32, ….is (a) – 1/16 (b) 16 (c) 1/16 (d) n...
Solution:
Given, T12 of the series –128, 64, –32, ….is to be determined.
We can observe that the series is a Geometric Progression (GP) with the first term (a) as –128 and the common ratio (r) as –1/2.
Using the formula for the nth term of a GP, we have:
Tn = a * r^(n-1)
Substituting the values, we get:
T12 = –128 * (-1/2)^(12-1)
T12 = –128 * (-1/2)^11
T12 = –128 * (-1/2048)
T12 = 1/16
Therefore, option (c) 1/16 is the correct answer.
HTML form:
Solution:
Given: T12 of the series –128, 64, –32, ….is to be determined.
Explanation:
- The given series is a Geometric Progression (GP) with the first term (a) as –128 and the common ratio (r) as –1/2.
- Using the formula for the nth term of a GP, we have: Tn = a * r^(n-1)
- Substituting the values, we get: T12 = –128 * (-1/2)^(12-1)
- Simplifying, we get: T12 = 1/16
Answer: Option (c) 1/16 is the correct answer.
Given, T12 of the series –128, 64, –32, ….is to be determined.
We can observe that the series is a Geometric Progression (GP) with the first term (a) as –128 and the common ratio (r) as –1/2.
Using the formula for the nth term of a GP, we have:
Tn = a * r^(n-1)
Substituting the values, we get:
T12 = –128 * (-1/2)^(12-1)
T12 = –128 * (-1/2)^11
T12 = –128 * (-1/2048)
T12 = 1/16
Therefore, option (c) 1/16 is the correct answer.
HTML form:
Solution:
Given: T12 of the series –128, 64, –32, ….is to be determined.
Explanation:
- The given series is a Geometric Progression (GP) with the first term (a) as –128 and the common ratio (r) as –1/2.
- Using the formula for the nth term of a GP, we have: Tn = a * r^(n-1)
- Substituting the values, we get: T12 = –128 * (-1/2)^(12-1)
- Simplifying, we get: T12 = 1/16
Answer: Option (c) 1/16 is the correct answer.
Community Answer
T12 of the series –128, 64, –32, ….is (a) – 1/16 (b) 16 (c) 1/16 (d) n...
C.1/16
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T12 of the series –128, 64, –32, ….is (a) – 1/16 (b) 16 (c) 1/16 (d) none of these?
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T12 of the series –128, 64, –32, ….is (a) – 1/16 (b) 16 (c) 1/16 (d) none of these? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about T12 of the series –128, 64, –32, ….is (a) – 1/16 (b) 16 (c) 1/16 (d) none of these? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for T12 of the series –128, 64, –32, ….is (a) – 1/16 (b) 16 (c) 1/16 (d) none of these?.
T12 of the series –128, 64, –32, ….is (a) – 1/16 (b) 16 (c) 1/16 (d) none of these? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about T12 of the series –128, 64, –32, ….is (a) – 1/16 (b) 16 (c) 1/16 (d) none of these? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for T12 of the series –128, 64, –32, ….is (a) – 1/16 (b) 16 (c) 1/16 (d) none of these?.
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