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Consider an infinite geometric series with first term a and common ratio r. if its sum is 4 and the second term is 3/4 ,then
- a)a = 4/7,r =3/7
- b)a = 2,r = 3/8
- c)a = 3/2 ,r = 1/2
- d)a = 3,r = 1/4
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
Consider an infinite geometric series with first term a and common rat...
2nd term of GP = a.r = 3 / 4
sum to infinity = 4
=> a / (1 - r) = 4
=> a = 4 - 4.r
=> a = 4 - 4 . (3 / 4a) [Since, a.r = 3/2]
=> a^2 - 4a + 3 = 0
=> a = 3 and a = 1
therefore for a=3 ; r =1/4
and for a = 1 ; r= 3/4
So, Option D is the correct answer
Community Answer
Consider an infinite geometric series with first term a and common rat...
Given Information:
- First term, a
- Common ratio, r
- Sum of the infinite geometric series, 4
- Second term, 3/4
Formula for the sum of an infinite geometric series:
S = a / (1 - r)
Calculating the values of a and r:
Given that the second term is 3/4, we can write:
ar = 3/4
Given that the sum of the infinite geometric series is 4, we can write:
4 = a / (1 - r)
Now, we can substitute the value of ar from the first equation into the second equation and solve for a and r.
ar = 3/4
a = 3/(4r)
4 = a / (1 - r)
4 = 3/(4r) / (1 - r)
4 = 3 / (4r - 4r^2)
4r^2 - 4r + 3 = 0
(4r - 3)(r - 1) = 0
r = 3/4 or r = 1
Since r cannot be equal to 1 (as it would make the series a constant series), we take r = 3/4.
Substituting r = 3/4 back into a = 3/(4r), we get:
a = 3 / (4 * 3/4) = 3/4
Therefore, the values of a and r are a = 3/4 and r = 3/4, which matches with option 'D'.
- First term, a
- Common ratio, r
- Sum of the infinite geometric series, 4
- Second term, 3/4
Formula for the sum of an infinite geometric series:
S = a / (1 - r)
Calculating the values of a and r:
Given that the second term is 3/4, we can write:
ar = 3/4
Given that the sum of the infinite geometric series is 4, we can write:
4 = a / (1 - r)
Now, we can substitute the value of ar from the first equation into the second equation and solve for a and r.
ar = 3/4
a = 3/(4r)
4 = a / (1 - r)
4 = 3/(4r) / (1 - r)
4 = 3 / (4r - 4r^2)
4r^2 - 4r + 3 = 0
(4r - 3)(r - 1) = 0
r = 3/4 or r = 1
Since r cannot be equal to 1 (as it would make the series a constant series), we take r = 3/4.
Substituting r = 3/4 back into a = 3/(4r), we get:
a = 3 / (4 * 3/4) = 3/4
Therefore, the values of a and r are a = 3/4 and r = 3/4, which matches with option 'D'.
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Consider an infinite geometric series with first term a and common ratio r. if its sum is 4 and the second term is 3/4 ,thena)a = 4/7,r =3/7b)a = 2,r = 3/8c)a = 3/2 ,r = 1/2d)a = 3,r = 1/4Correct answer is option 'D'. Can you explain this answer?
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Consider an infinite geometric series with first term a and common ratio r. if its sum is 4 and the second term is 3/4 ,thena)a = 4/7,r =3/7b)a = 2,r = 3/8c)a = 3/2 ,r = 1/2d)a = 3,r = 1/4Correct answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Consider an infinite geometric series with first term a and common ratio r. if its sum is 4 and the second term is 3/4 ,thena)a = 4/7,r =3/7b)a = 2,r = 3/8c)a = 3/2 ,r = 1/2d)a = 3,r = 1/4Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider an infinite geometric series with first term a and common ratio r. if its sum is 4 and the second term is 3/4 ,thena)a = 4/7,r =3/7b)a = 2,r = 3/8c)a = 3/2 ,r = 1/2d)a = 3,r = 1/4Correct answer is option 'D'. Can you explain this answer?.
Consider an infinite geometric series with first term a and common ratio r. if its sum is 4 and the second term is 3/4 ,thena)a = 4/7,r =3/7b)a = 2,r = 3/8c)a = 3/2 ,r = 1/2d)a = 3,r = 1/4Correct answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Consider an infinite geometric series with first term a and common ratio r. if its sum is 4 and the second term is 3/4 ,thena)a = 4/7,r =3/7b)a = 2,r = 3/8c)a = 3/2 ,r = 1/2d)a = 3,r = 1/4Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider an infinite geometric series with first term a and common ratio r. if its sum is 4 and the second term is 3/4 ,thena)a = 4/7,r =3/7b)a = 2,r = 3/8c)a = 3/2 ,r = 1/2d)a = 3,r = 1/4Correct answer is option 'D'. Can you explain this answer?.
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Consider an infinite geometric series with first term a and common ratio r. if its sum is 4 and the second term is 3/4 ,thena)a = 4/7,r =3/7b)a = 2,r = 3/8c)a = 3/2 ,r = 1/2d)a = 3,r = 1/4Correct answer is option 'D'. Can you explain this answer?, a detailed solution for Consider an infinite geometric series with first term a and common ratio r. if its sum is 4 and the second term is 3/4 ,thena)a = 4/7,r =3/7b)a = 2,r = 3/8c)a = 3/2 ,r = 1/2d)a = 3,r = 1/4Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of Consider an infinite geometric series with first term a and common ratio r. if its sum is 4 and the second term is 3/4 ,thena)a = 4/7,r =3/7b)a = 2,r = 3/8c)a = 3/2 ,r = 1/2d)a = 3,r = 1/4Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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