If a number a^2, b^2, c^2 are in A.P., show that 1/b+c, 1/c+a, 1/a+b are in A.P.
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Answer:
If a^2 b^2 c^2 are in AP, does that prove that 1/b+c, 1/c+a, 1/a+b are in AP?
If a^2 , b^2 and c^2 are in AP, then what is a/(b+c),b/(c+a) c/(a+b)?
Is this question true- if a^2, b^2 ,c^2 are in AP then prove that a/b+c, b/c+a, c/a+b are in AP?
How can I prove that if “a^2 (b+c), b^2(c+a) and c^2 (a+b) are in AP then either a, b, c are in AP or ab+bc+ca=0” ?
If 1b+c,1c+a,1a+b are in AP, then what are a2,b2 and c2 in?
There is a short way.
a2 , b2 , c2 in AP , implies that,
a2+ab+bc+ca , b2+ab+bc+ca , c2+ab+bc+ca are in AP, implies that [Adding ab+bc + ca to every term of the above terms. This is allowed in an AP.]
(a+b).(a+c), (b+a).(a+c), (c+a).(c+b) are in AP [By factorising each term of AP, as in above]
Thus, 1b+c , 1c+a , 1a+b are in AP [By dividing the above factorised terms , by (a+b).(b+c).(c+a). This is allowed in an AP.]
QED.
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If a^2 ,b^2. and c^2. are in A.P.then prove that 1/(b+c),1/(c+a),1/(a+b) are in A.P.
a^2. , b^2. , c^2. are in A.P.
by adding ab+b.c+c.a. in. each. term.
a^2+ab+b.c+ca. , b^2+a.b+b.c+c.a. , c^2+a.b+b.c+c.a. are in A.P.
(a+b)(c+a). , (a+b) (b+c). , (b+c)(c+a). are in A.P.
On dividing by (a+b)(b+c)(c+a) in each term
or. 1/(b+c). , 1/(c+a). , 1/(a+b). are in A.P. , proved.
If a^2 b^2 c^2 are in AP, does that prove that 1/b+c, 1/c+a, 1/a+b are in AP?
If a^2 , b^2 and c^2 are in AP, then what is a/(b+c),b/(c+a) c/(a+b)?
Is this question true- if a^2, b^2 ,c^2 are in AP then prove that a/b+c, b/c+a, c/a+b are in AP?
How can I prove that if “a^2 (b+c), b^2(c+a) and c^2 (a+b) are in AP then either a, b, c are in AP or ab+bc+ca=0” ?
If 1b+c,1c+a,1a+b are in AP, then what are a2,b2 and c2 in?
If a+b+c ≠ 0 and a/(b+c), b/(c+a), c/(b+a) are in AP, how do I prove that 1/(b+c), 1/(c+a), 1/(a+b) are also in AP?
If 1/(a+b), 1/2b, 1/(b+c) are in AP, how do you prove that a, b, c are in GP?
How do I prove that 1a(b+1)+1b(c+1)+1c(d+1)+1d(a+1)≥2 if abcd=1 where a,b,c,d are positive reals?
If a+b+c+d = 2, then what is the value of (1+a)(1+b)(1+c)(1+d)?
If a b c are in GP prove that (a^2+ab+b^2) / (bc+CA+ab) =(b+a) /(c+b)?
If a2 , b2 , and c2 are in AP, how can I show that b+c, c+a, a+b are in HP?
If a+b+c=0 , then what is 1b2+c2−a2+1c2+a2−b2+1a2+b2−c2 equal to?
How can I prove that:-bc(b^3-c^3) +ca (c^3-a^3) +ab (a^3-b^3) =(b-c) (c-a) (a-b) (a^2+b^2+c^2+ab+bc+ca)?
How do I prove b1d1<b2d2 implies с1+b1c2+d1<с1+b2c2+d2(c1,b1,b2∈R;c2,d1,d2>0)?
How do we show that if (a,b) = 1 then (a+b, a-b)=1 or 2?
Step-by-step explanation:
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