Question

Suppose a,b,c are in A.P and $a^2$,$b^2$,$c^2$ are in GP. If a < b < c and a+b+c = $\frac{3}{2}$, then the value of a is

• A. +
• B. -
• C. -
• D. +

Answer 1

The correct option is B

-

Let the number be

m - d, m, m + d
$\downarrow$ $\downarrow$ $\downarrow$
a b c

We will use the given three condition to find the value of m and d.

The first condition that they are in A.P is used to form the A.P.

There is one more condition (a<b<c)given,

Which we will use to eliminate any values of d and m at the end.

a+b+c = $\frac{3}{2}$

$\Rightarrow$ m-d + m + m + d = $\frac{3}{2}$

$\Rightarrow$ 3m = $\frac{3}{2}$

m = $\frac{1}{2}$

$(m-d{)}^2$,$(m{)}^2$ , $(m+d{)}^2$ are in GP

$\Rightarrow$ $(m+d{)}^2$ $(m-d{)}^2$ = $(m{)}^4$

$(m^2-d^2{)}^2$ = $(m{)}^4$

$m^2-d^2$ = $\overline{+}$ $m^2$

$d^2$ = $m^2$ $\overline{+}$ $m^2$

$d^2$ = 2$m^2$ or $d^2$ = 0

d ≠ 0 since a,b,c are different number (given a<b<c)

$\Rightarrow$ d = $\overline{+}$ $\sqrt{2}$m

= $\overline{+}$ $\sqrt{2}$ $\times$ $\frac{1}{2}$

= $\overline{+}$ $\frac{1}{\sqrt{2}}$

Since a<b<c,d is positive

$\Rightarrow$ d = $\overline{+}$ $\frac{1}{\sqrt{2}}$

$\Rightarrow$ a = m - d

= $\frac{1}{2}$ - $\frac{1}{\sqrt{2}}$