# Let a,b,c>0 and a,b,c are in A.P. a^2,b^2,c^2 are in G.P. then choose the correct one ? (a)a=b=c, (b)a^2+b^2=c^2 , (c) a^2+c^2=3 b^2, (d) none of these

Sep 28, 2017

$a = b = c$

#### Explanation:

The generic terms of an AP sequence can be represented by:

$\textsf{\left\{a , a + d , a + 2 d\right\}}$

We are told that $\left\{a , b , c\right\}$, and we note that if we take a higher term and subtract its previous term we get the common difference; thus

$c - b = b - a$
$\therefore 2 b = a + c$ ..... [A]

The generic terms of an GP sequence can be represented by:

$\textsf{\left\{a , a r , a {r}^{2}\right\}}$

We are told that $\left\{{a}^{2} , {b}^{2} , {c}^{2}\right\}$, and we note that if we take a higher term and divide by its previous term we get the common ratio, thus:

${c}^{2} / {b}^{2} = {b}^{2} / {a}^{2} \implies \frac{c}{b} = \frac{b}{a} \setminus \setminus \setminus \setminus$ (as $a , b , c > 0$)
$\therefore {b}^{2} = a c$ ..... [B]

Substituting [A] into [B] we have:

${\left(\frac{a + c}{2}\right)}^{2} = a c$
$\therefore {a}^{2} + 2 a c + {c}^{2} = 4 a c$
$\therefore {a}^{2} - 2 a c + {c}^{2} = 0$
$\therefore {\left(a - c\right)}^{2} = 0$
$\therefore a = c$

And if we substitute $a = c$ into Eq [B], we have:

${b}^{2} = {c}^{2} \implies b = c \setminus \setminus \setminus \setminus$ (as $a , b , c > 0$)

Hence we have $a = c$ and $b = c \implies a = b = c$