Suppose four distinct positive numbers a_1, a | vedclass

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(C) Let the $G.P.$ be $a, ar, ar^2, ar^3$ where $r 
eq 1$ and $r > 0$.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,

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$STATEMENT-1$ is True,$STATEMENT-2$ is False

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(C) Let the $G.P.$ be $a, ar, ar^2, ar^3$ where $r 
eq 1$ and $r > 0$.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 $A.P.$,$b_2-b_1 = b_3-b_2 = b_4-b_3$ must hold.$b_2-b_1 = ar$,$b_3-b_2 = ar^2$,$b_4-b_3 = ar^3$.Since $r 
eq 1$ and $r > 0$,$ar 
eq ar^2 
eq ar^3$,so they are not in $A.P.$For $G.P.$,$\frac{b_2}{b_1} = 1+r$ and $\frac{b_3}{b_2} = \frac{1+r+r^2}{1+r}$. These are not equal for $r 
eq 0, 1$.For $H.P.$,the reciprocals must be in $A.P.$,which is clearly not true.Thus,$STATEMENT-1$ is True and $STATEMENT-2$ is False.

Suppose four distinct positive numbers $a_1, a_2, a_3, a_4$ are in $G.P.$ Let $b_1=a_1, b_2=b_1+a_2, b_3=b_2+a_3$ and $b_4=b_3+a_4$.
$STATEMENT-1$ : The numbers $b_1, b_2, b_3, b_4$ are neither in $A.P.$ nor in $G.P.$
$STATEMENT-2$ : The numbers $b_1, b_2, b_3, b_4$ are in $H.P.$

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Suppose four distinct positive numbers $a_1, a_2, a_3, a_4$ are in $G.P.$ Let $b_1=a_1, b_2=b_1+a_2, b_3=b_2+a_3$ and $b_4=b_3+a_4$.$STATEMENT-1$ : The numbers $b_1, b_2, b_3, b_4$ are neither in $A.P.$ nor in $G.P.$$STATEMENT-2$ : The numbers $b_1, b_2, b_3, b_4$ are in $H.P.$