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If $a,\,b,\,c$ are in $A.P.$ and ${a^2},\,{b^2},{c^2}$ are in $H.P.$, then
$a \ne b \ne c$
${a^2} = {b^2} = \frac{{{c^2}}}{2}$
$a,\,b,\,c$ are in $G.P.$
$\frac{{ - a}}{2},b,c$ are in $G.P$
Solution
(d) $a, b, c$, are in $A.P.$
$⇒$ $2b = a + c,b -a = c -b$
${a^2},{b^2},{c^2}$ are in $H.P.$
$\frac{1}{{{b^2}}} – \frac{1}{{{a^2}}} = \frac{1}{{{c^2}}} – \frac{1}{{{b^2}}}$ $ \Rightarrow \frac{{{a^2} – {b^2}}}{{{a^2}{b^2}}} = \frac{{{b^2} – {c^2}}}{{{b^2}{c^2}}}$
$⇒$ $(a – b)[{c^2}(a + b) – {a^2}(b + c)] = 0$,
$[\because \,(b – c) = (a – b)]$
$⇒$ $a = b$ or ${c^2}a + {c^2}b – {a^2}b – {a^2}c = 0$
$⇒$ ${c^2}a + {c^2}b – {a^2}b – {a^2}c = 0$
$⇒$ $ac\,(c – a) = b\,({a^2} – {c^2})$
$⇒$ $ac = – b\,(c + a)$
$⇒$ $ – ac = b.2b$
$⇒$ ${b^2} = ( – a/2)\,c$,
$\therefore – a/2,b,c$ are in $G.P.$
