If `A_1, A_2, G_1, G_2, ; a n dH_1, H_2` are two arithmetic, geometric and harmonic means respectively, between two quantities `aa n db ,t h e na b` is equal to `A_1H_2` b. `A_2H_1` c. `G_1G_2` d. none of these

Options

A. `A_H_2`
B. `A_2H_1`
C. `G_1G_2`
D. none of these

Answer

A, B, C

Explanation

Since `A_(1),A_(2)` are two arthimatic means between a and b, therefore, `a,A_(1),A_(2)b` are in A.P. with common difference d given by
`d=(b-a)/(2+1)=(b-a)/3` [using `d=(b-a)/(n+1)`]
Now,`A_(1)=a+d=a+(b-a)/3=(2a+b)/3`
and `A_(2)=a+2d=a+2((b-a)/3)=(a+2b)/3`
Iti s given that `G_(1),G_(2)` are two geometric means between a and b. Therefore,a,`G_(1),G_(2),b` are in G.P. with common ratio r given by
`r=(b/a)^(1/(2+1))=(b/a)^(1//3)` `[becauser=(b/a)^(1/(n+1))]`
Now, `G_(1)=ar=a(b/a)^(1//3)=a^(2//3)b^(1//3)`
and `G_(2)=ar^(2)=a(b/a)^(2//3)=a^(1//3)b^(2//3)`
It is also given that `H_(1),H_(2)` are two harmonic means between a and b, therefore,a `H_(1),H_(2),` b are in H.P. Hence, `1//a,1//H_(1),1//H_(2),1//b`, are in A.P. with common difference D given by
`D=(a-b)/((2+1)ab)=(a-b)/(3ab)` `[becauseD=(a-b)/((n+1)ab)]`
Now, `1/H_(1)=1/a+D=1/a+(a-b)/(3ab)=(a+2b)/(3ab)`
or `H_(1)=(3ab)/(a+2b)`
`1/(H_(2))=1/a+2D`
`=1/a+(2(a-b))/(3ab)`
`=(2a+b)/(3ab)` ltbr or `H_(2)=(3ab)/(2a+b)`
we have,
`A_(1)H_(2)=(2a+b)/3xx(3ab)/(2a+b)=ab`,
`A_(2)H_(1)=(a+2b)/3xx(3ab)/(a+2b)=ab`,
`G_(1)G_(2)=(a^(2//3)b^(1//3))(a^(1//3)b^(2//3))=ab`
`thereforeA_(1)H_(2)=A_(2)H_(1)=G_(1)G_(2)==ab`