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STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
MCQ
If $a,\,b,\,c$ are in $A.P.$ and ${a^2},\,{b^2},{c^2}$ are in $H.P.$, then
  • A
    $a \ne b \ne c$
  • B
    ${a^2} = {b^2} = \frac{{{c^2}}}{2}$
  • C
    $a,\,b,\,c$ are in $G.P.$
  • ✓
    $\frac{{ - a}}{2},b,c$are in $G.P$
✓

Answer

Correct option: D.
$\frac{{ - a}}{2},b,c$are in $G.P$
d
(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 $G.P.$

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