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where a, b, c are positive integers then a+b+c=
Answers
Answer:
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Answer:
Given a, b and c are positive integers and
1/(a+ (1/(b+ (1/ (c+1/2))))) = 16/23 ……………………………………………………………….(1)
Or, (a+ (1/(b+ (1/ (c+1/2))))) = 23/16 [Taking reciprocals on both sides]
Or, a+ (1/(b+ (1/ (c+1/2)))) = 1 + 7/16
Since a is a positive integer and the right-side consists of a positive integer and a fraction,
a = 1
∴ (1/(b+ (1/ (c+1/2)))) = 7/16
Again taking reciprocals on both sides,
(b+ (1/ (c+1/2))) = 16/7
Or, b+ (1/ (c+1/2)) = 2 + 2/7
By the same argument as before for a,
b = 2
∴ 1/ (c+1/2) = 2/7
Taking reciprocal for the last time,
c + 1/2 = 7/2
Or, c = 7/2 - 1/2 = (7–1)/2 = 6/2 = 3
Thus a = 1, b = 2, c = 3
and the mean of a, b and c
= (a+b+c)/3 = (1+2+3)/3
= 6/3
= 2
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· Answer requested by Yashwant Choudhary
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Other Answers

Nancy Mitchell, used to be a teacher.
Answered 2 years ago · Author has 2.6K answers and 3.4M answer views
What is the value of a, b, and c if 1/a+1/b+1/c=9/26?
What is the value of a, b and c if 1/a+1/b+1/c=9/26?
The problem does not specify that aa, bb and cc have to be integers.
The formula
1N=1N+1+1N(N+1)1N=1N+1+1N(N+1)
expresses a unit fraction as the sum of two other unit fractions. A derivation of this formula is available online.
Letting N=26 N=26
⟹126=126+1+126(26+1)⟹126=126+1+126(26+1)
=127+126(27)