Question

Suppose four distinct positive numbers $a_1,a_2,a_3$ and $a_4$ are in $G.P$. Let $b_1=a_1,b_2=b_1+a_2,b_3=b_2+a_3$ and $b_4=b_3+a_4,$ then $b_1,b_2,b_3$ and $b_4$ are in

• A. A.P.
• B. G.P.
• C. H.P.
• D. none of these

Answer 1

Answer: (D)

Here, $b_1=a_1$
$b_2=a_1+a_2=a_1(1+r)$
$b_3=a_1(1+r)+a_1{r}^2=a_1\left(1+r+r^2\right)$
$b_4=a_1\left(1+r+r^2\right)+a_1{r}^3=a_1\left(1+r+r^2+r^3\right),r$ being the common ratio of the $G.P$.
Clearly, $b_1,b_2,b_3$ and $b_4$ are in neither of $A.P.,G.P$. and $H.P$.