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If 'n' arithmetic means are inserted between 7 & 71 and 5th arithmetic mean is 27, then 'n' is equal to:
- a)15
- b)16
- c)17
- d)18
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
If n arithmetic means are inserted between 7 & 71 and 5tharithmeti...
And 37, what is the common difference between the arithmetic means?
There are n+1 terms in the sequence (including 7 and 37), and the common difference between consecutive terms is d.
Using the formula for the nth term of an arithmetic sequence, we can write:
37 = 7 + (n+1)d
Simplifying this equation, we get:
30 = nd
So the number of arithmetic means, n, is equal to 30/d.
To find the common difference between the arithmetic means, we can use the formula for the common difference in an arithmetic sequence:
d = (37 - 7)/(n+1)
Substituting 30/d for n, we get:
d = (37 - 7)/(30/d + 1)
Multiplying both sides by (30/d + 1), we get:
d(30/d + 1) = 30
Expanding and simplifying, we get:
d^2 + d - 30 = 0
This is a quadratic equation that can be factored as:
(d + 6)(d - 5) = 0
So the possible values for d are -6 and 5. However, since we are looking for a positive common difference between the arithmetic means, we can discard the negative solution and conclude that the common difference is 5.
There are n+1 terms in the sequence (including 7 and 37), and the common difference between consecutive terms is d.
Using the formula for the nth term of an arithmetic sequence, we can write:
37 = 7 + (n+1)d
Simplifying this equation, we get:
30 = nd
So the number of arithmetic means, n, is equal to 30/d.
To find the common difference between the arithmetic means, we can use the formula for the common difference in an arithmetic sequence:
d = (37 - 7)/(n+1)
Substituting 30/d for n, we get:
d = (37 - 7)/(30/d + 1)
Multiplying both sides by (30/d + 1), we get:
d(30/d + 1) = 30
Expanding and simplifying, we get:
d^2 + d - 30 = 0
This is a quadratic equation that can be factored as:
(d + 6)(d - 5) = 0
So the possible values for d are -6 and 5. However, since we are looking for a positive common difference between the arithmetic means, we can discard the negative solution and conclude that the common difference is 5.
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If n arithmetic means are inserted between 7 & 71 and 5tharithmeti...
A) 16
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If n arithmetic means are inserted between 7 & 71 and 5tharithmetic mean is 27, then n is equal to:a)15b)16c)17d)18Correct answer is option 'A'. Can you explain this answer?
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