CA Foundation Exam > CA Foundation Questions > If a, b, c are in G.P. then a2+b2, ab+bc, b2+...
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If a, b, c are in G.P. then a2+b2, ab+bc, b2+c2 are in
- a)A.P.
- b)G.P.
- c)H.P.
- d)None
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
If a, b, c are in G.P. then a2+b2, ab+bc, b2+c2are ina)A.P.b)G.P.c)H.P...
As we are given a, b,c are in G.P:
lets assume, a=b/r; b=b; c=br
Where r → common ratio between the terms such that ratio of c/b=b/a
Now, we calculate value of a^2 + b^2, ab+ bc, b^2 + c^2 to prove that these are in G.P
1st term = a^2 + b^2 = (b/r)^2 + b^2 = b^2[(1+r^2)/r^2]
2nd term: ab + bc = b/r x b + b x br = b^2[(1+r^2)/r]
3rd term: b^2 + c^2 = b^2 + b^2 x r^2 = b^2(1+r^2)
3rd term/2nd term = [b^2(1+r^2)]/b^2[(1+r^2)/r] = b^2(1+r^2) x r/b^2(1+r^2) = r …(1)
2nd term/1st term = b^2(1+r^2)/r x r^2/b^2(1+r^2) = r …(2)
From eq (1) & (2)
common ratios are same so a^2 + b^2, ab+ bc, b^2 + c^2 in G.P
Most Upvoted Answer
If a, b, c are in G.P. then a2+b2, ab+bc, b2+c2are ina)A.P.b)G.P.c)H.P...
Given, a, b, c are in G.P.
So, b/a = c/b
b^2 = ac
Multiplying both sides by a and c, we get:
a^2 c^2 = a^3 c.b = a^2 b^2 c^2
a^2 b^2 c^2 is common in above two equations.
So, a^3 c.b = a^2 b^2 c^2
On simplifying, we get:
a/b = b/c = a^2 b^2 c^2 / a^3 c.b
Now, let's calculate the product of the given terms:
(a^2 b^2) (b c) = a^2 b^2 c^2
This is the common ratio of the given terms.
Hence, the given terms are in G.P.
So, b/a = c/b
b^2 = ac
Multiplying both sides by a and c, we get:
a^2 c^2 = a^3 c.b = a^2 b^2 c^2
a^2 b^2 c^2 is common in above two equations.
So, a^3 c.b = a^2 b^2 c^2
On simplifying, we get:
a/b = b/c = a^2 b^2 c^2 / a^3 c.b
Now, let's calculate the product of the given terms:
(a^2 b^2) (b c) = a^2 b^2 c^2
This is the common ratio of the given terms.
Hence, the given terms are in G.P.
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If a, b, c are in G.P. then a2+b2, ab+bc, b2+c2are ina)A.P.b)G.P.c)H.P.d)NoneCorrect answer is option 'B'. Can you explain this answer?
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