Let four distinct positive numbers a_1, a_2, | vedclass

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(C) Let the geometric progression be $a, ar, ar^2, ar^3$ where $a > 0$ and $r > 0, r 
eq 1$ (since numbers are distinct).Then $b_1 = a$,$b_2 = a(1+r)

Vedclass · Accepted Answer

Statement-$I$ is true. Statement-$II$ is false.

Vedclass · Answer

(C) Let the geometric progression be $a, ar, ar^2, ar^3$ where $a > 0$ and $r > 0, r 
eq 1$ (since numbers are distinct).Then $b_1 = a$,$b_2 = a(1+r)$,$b_3 = a(1+r+r^2)$,$b_4 = a(1+r+r^2+r^3)$.For $b_1, b_2, b_3, b_4$ to be in $AP$,$b_2 - b_1 = b_3 - b_2$,which implies $ar = ar^2$,so $r=1$,which contradicts the distinctness.For $b_1, b_2, b_3, b_4$ to be in $GP$,$b_2^2 = b_1 b_3$,which implies $a^2(1+r)^2 = a^2(1+r+r^2)$,so $1+2r+r^2 = 1+r+r^2$,implying $r=0$,which is not possible for positive numbers.Thus,Statement-$I$ is true.For $HP$,the reciprocals must be in $AP$. $\frac{1}{b_1}, \frac{1}{b_2}, \frac{1}{b_3}, \frac{1}{b_4}$ are not in $AP$ for general $r$. Thus,Statement-$II$ is false.

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