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The three numbers between 1 and 256 such that the sequence is in GP are
- a)4,16,64
- b)2,8,32
- c)8,16,32
- d)8,16,64
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The three numbers between 1 and 256 such that the sequence is in GP ar...
nth G.M. between a and b is
Gn = arn
Where common ratio is r = (b/a)(1/(n+1))
So, to insert 3 geometric means between 1 and 256
r = (b/a)(1/(n+1))
r = (256/(1)(1/(3+1))
r = (256)1/4
r = (4)(4)1/4
r = 4
Gn = 1 * (4)n
G1 = 1 * (4)1 = 4
G2 = 1 * (4)2 = 16
G3 = 1 * (4)3 = 64
The terms are 4, 16, 64
Gn = arn
Where common ratio is r = (b/a)(1/(n+1))
So, to insert 3 geometric means between 1 and 256
r = (b/a)(1/(n+1))
r = (256/(1)(1/(3+1))
r = (256)1/4
r = (4)(4)1/4
r = 4
Gn = 1 * (4)n
G1 = 1 * (4)1 = 4
G2 = 1 * (4)2 = 16
G3 = 1 * (4)3 = 64
The terms are 4, 16, 64
Most Upvoted Answer
The three numbers between 1 and 256 such that the sequence is in GP ar...
According to the question first term is 1 and last is 256 the formula of insertation of numbers in gp is ar(b/a)power 1/n+1 so..... its second term will be 4 and third will be 16 by ar square and 64 by ar cube
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Community Answer
The three numbers between 1 and 256 such that the sequence is in GP ar...
Consider three no. as a/r,a,ar
Now,1*r^4=256
So,r=4
Again,1*r^2=a
So,a=16
Hence,your answer comes as 4,16,64
Now,1*r^4=256
So,r=4
Again,1*r^2=a
So,a=16
Hence,your answer comes as 4,16,64
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The three numbers between 1 and 256 such that the sequence is in GP area)4,16,64b)2,8,32c)8,16,32d)8,16,64Correct answer is option 'A'. Can you explain this answer?
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