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Question: a, b, c are in A.P. and the points A(a,1), B(b,2) and C(c.3) are such that\({\left( {OA} \right)^2},...

a, b, c are in A.P. and the points A(a,1), B(b,2) and C(c.3) are such that(OA)2,(OB)2 and (OC)2{\left( {OA} \right)^2},{(OB)^2}{\text{ and (OC}}{{\text{)}}^2} are also in A.P.;O being the origin, then
A. a2+c2=2b2−2 B. ac=b2+1 C. (a+c)2=4b2 D. a+b+c=3b  {\text{A}}{\text{. }}{a^2} + {c^2} = 2{b^2} - 2 \\\ {\text{B}}{\text{. }}ac = {b^2} + 1 \\\ {\text{C}}{\text{. (}}a + c{{\text{)}}^2} = 4{b^2} \\\ {\text{D}}{\text{. }}a + b + c = 3b \\\

Explanation

Solution

Hint: In this question, first we need to establish relation among a, b, c using the property of A.P. that common difference is the same. Similarly, we have to create relations between (OA)2,(OB)2 and (OC)2{\left( {OA} \right)^2},{(OB)^2}{\text{ and (OC}}{{\text{)}}^2} using property of A.P. Then, using the obtained equation we have to create a new relation which satisfies both the given conditions.

Complete step-by-step answer:
We have,
a, b, c are in A.P.
Then,
b-a=c-b
2b = a + c ------ eq.1
And, (OA)2,(OB)2 and (OC)2{\left( {OA} \right)^2},{(OB)^2}{\text{ and (OC}}{{\text{)}}^2} are also in A.P.
Then,
(OB)2−(OA)2=(OC)2−(OB)2{(OB)^2}-{(OA)^2}={(OC)^2}-{(OB)^2}
2(OB)2=(OA)2+(OC)22{(OB)^2} = {(OA)^2} + {(OC)^2} ----- eq.2
Now, using distance formula: distance between two given points (x,y) and (z,w)(x,y){\text{ and }}(z,w) is given by
distance = (z−x)2+(w−y)2{\text{distance = }}\sqrt {{{(z - x)}^2} + {{(w - y)}^2}}
We know, the coordinates of origin are (0,0).

⇒(OA)2=(a−0)2+(1−0)2 ⇒(OA)2=a2+1 —- eq.3  \Rightarrow {(OA)^2} = {\left( {a - 0} \right)^2} + {(1 - 0)^2} \\\ \Rightarrow {(OA)^2} = {a^2} + 1{\text{ ---- eq}}{\text{.3}} \\\

Similarly,
⇒(OB)2=(b−0)2+(2−0)2 ⇒(OB)2=b2+4 —–eq.4  \Rightarrow {(OB)^2} = {\left( {b - 0} \right)^2} + {(2 - 0)^2} \\\ \Rightarrow {(OB)^2} = {b^2} + 4{\text{ -----eq}}{\text{.4}} \\\

And,
⇒(OC)2=(c−0)2+(3−0)2 ⇒(OC)2=c2+9 ——eq.5  \Rightarrow {(OC)^2} = {\left( {c - 0} \right)^2} + {(3 - 0)^2} \\\ \Rightarrow {(OC)^2} = {c^2} + 9{\text{ ------eq}}{\text{.5}} \\\

Now, put values of (OA)2,(OB)2 and (OC)2{\left( {OA} \right)^2},{(OB)^2}{\text{ and (OC}}{{\text{)}}^2} from eq.3 ,eq.4, eq.5 in eq.2 we get
⇒2(b2+4)=a2+1+c2+9\Rightarrow 2({b^2} + 4) = {a^2} + 1 + {c^2} + 9
On Solving above equation, we get
⇒2b2=a2+c2+2 ⇒a2+c2=2b2−2 —- eq.6  \Rightarrow 2{b^2} = {a^2} + {c^2} + 2 \\\ \Rightarrow {a^2} + {c^2} = 2{b^2} - 2{\text{ ---- eq}}{\text{.6}} \\\
Therefore, option A. is correct.
We know,
(x+y)2=x2+2xy+y2{(x + y)^2} = {x^2} + 2xy + {y^2}
⇒x2+y2=(x+y)2−2xy\Rightarrow {x^2} + {y^2} = {(x + y)^2} - 2xy
Using above identity, we can rewrite eq.6 as
⇒(a+c)2−2ac=2b2−2 —–eq.7  ∴(a + c)2−2ac=a2+c2\Rightarrow {(a + c)^2} - 2ac = 2{b^2} - 2{\text{ -----eq}}{\text{.7 \\{ }}\therefore {{\text{(a + c)}}^2} - 2ac = {a^2} + {c^2}\\}
From eq. 1 we can rewrite above equation as
⇒(2b)2−2ac=2b2−2\Rightarrow {(2b)^2} - 2ac = 2{b^2} - 2
On solving above equation, we get
⇒4b2−2b2+2=2ac ⇒ac=b2+1 ——-eq.8  \Rightarrow 4{b^2} - 2{b^2} + 2 = 2ac \\\ \Rightarrow ac = {b^2} + 1{\text{ -------eq}}{\text{.8}} \\\
Therefore, option B. is correct
Put value of ac from eq.8 into eq.7, we get
⇒(a+c)2−2(b2+1)=2b2−2\Rightarrow {(a + c)^2} - 2({b^2} + 1) = 2{b^2} - 2
On solving above equation, we get
⇒(a+c)2=4b2\Rightarrow {(a + c)^2} = 4{b^2}
Therefore, option C. is correct.
⇒a+c=2b\Rightarrow a + c = 2b
On adding b to both sides in above equation, we get
⇒a+b+c = 3b\Rightarrow a + b + c{\text{ = 3}}b
Therefore, option D is correct.
Hence, option A, B, C, D are correct answers.

Note:- Whenever you get this type of question the key concept to solve is to learn the concept of Arithmetic Mean when given three numbers are in A.P. When three numbers a, b, c are in Arithmetic Progression(A.P.) then b is called the arithmetic mean of the numbers “a” and “c’. And the difference between two consecutive numbers in A.P. remains constant.