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Sequences & Series: Between Between

JEE Maths question with a full step-by-step solution.

Question

a, b, c

are in A.P.,

p, p'

are the A.M. and G.M. between

a

and

b

, and

q, q'

are the A.M. and G.M. between

b

and

c

, then

p^2+q^2=p'^2+q'^2

pq=p'q'

p^2-q^2=p'^2-q'^2

correct

p^2+p'^2=q^2+q'^2

Solution

Step 1: Here

p=\dfrac{a+b}{2}

q=\dfrac{b+c}{2}

p'=\sqrt{ab}

q'=\sqrt{bc}

, with

2b=a+c.

Step 2:

p^2-q^2=\frac{(a+b)^2-(b+c)^2}{4}=\frac{(a-c)(a+2b+c)}{4}=\frac{(a-c)\cdot 4b}{4}=b(a-c),

using

a+c=2b

a+2b+c=4b.

Step 3:

p'^2-q'^2=ab-bc=b(a-c).

p^2-q^2=p'^2-q'^2.

Correct answer: (3)

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