Question

If between two numbers one geometric mean is G and two arithmetic means are p and q, prove that : G² = (2p-q) (2q-p).

Please give the correct solution.

Answer 1

g = (ab)1/2  is quite simple.now a , p , q, b are in AP .b =  a +(4-1) d  ;   d = b -a/3thereofore ,  p =  a +(b-a/3) =  2a+b/3                     q  = p +d =  (2a+b/3) + (b -a)/3 = 2b+a/3so 2p-q  = 3a/3 =a  and p -2q  = -bso , (2p-q)(p-2q) =  -ab =  -g^2

Answer 2

Answer:  The proof is done below.

Step-by-step explanation:  We are given that in between two numbers, one geometric mean is G and two arithmetic means are p and q.

We are to prove that :

$G^2=(2p-q)(2q-p).$

Let a and b be the two numbers.

Then, according to the given information, we have

a, G, q are in G.P. (geometric progression).

So,

$\dfrac{G}{a}=\dfrac{b}{G}\\\\\Rightarrow G^2=ab~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

and

a, p, q, b are in A.P. (arithmetic progression).

So,

$p-a=q-p\\\\\Rightarrow a=p-q+p\\\\\Rightarrow a=p-q~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)\\\\q-p=b-q\\\\\Rightarrow b=q-p+q\\\\\Rightarrow b=q-p~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)$

Substituting the values of a and b from equations (ii) and (iii) in equation (i), we get

$G^2=ab\\\\\Rightarrow G^2=(2p-q)(2q-p).$

Hence proved.