Question

Assertion :Suppose four distinct positive numbers $a_1,a_2,a_3,a_4$ are in GP. 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 AP nor in GP, and Reason: STATEMENT- 1: The numbers $b_1,b_2,b_3,b_4$ are in HP.

• A. STATEMENT- 1 is True, STATEMENT- 2 is True; STATEMENT- 2 is a correct explanation for STATEMENT- 1.
• B. STATEMENT- 1 is True, STATEMENT- 2 is True; STATEMENT- 2 is NOT a correct explanation for STATEMENT- 1.
• C. STATEMENT- 1 is True, STATEMENT- 2 is False.
• D. STATEMENT- 1 is False, STATEMENT- 2 is True.

Answer 1

The correct option is C STATEMENT- 1 is True, STATEMENT- 2 is False.

$\because a_1,a_2,a_3$ and $a_4$are in G.P with common ratio ${}^{\prime }{r}^{\prime }$
$\Rightarrow a_1=a,a_2=ar,a_3=a{r}^{2}and{a}_4=a{r}^3$
$\because b_1=a_1=a$
$b_2=b_1+a_2=a+ar=a(1+r)$
$b_3=b_2+a_3=a(1+r)+a{r}^2=a(1+r+r^2)$
$b_4=b_3+a_4=a(1+r+r^2)+a{r}^3=a(1+r+r^2+r^3)$
$\Rightarrow$ Option (c) is correct