If a,,b,,c are in A.P. and {a^2},,{b^2},{c^2} | vedclass

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Question

(d) $a, b, c$, are in $A.P.$

$&rArr;$  $2b = a + c,b -a = c -b$

${a^2},{b^2},{c^2}$ are in $H.P.$

$\frac{1}{{{b^2}}} - \frac{1}{{{a^2}}}

Vedclass · Accepted Answer

$\frac{{ - a}}{2},b,c$ are in $G.P$

Vedclass · Answer

(d) $a, b, c$, are in $A.P.$

$&rArr;$  $2b = a + c,b -a = c -b$

${a^2},{b^2},{c^2}$ 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}}}$

$&rArr;$  $(a - b)[{c^2}(a + b) - {a^2}(b + c)] = 0$,

$[\because \,(b - c) = (a - b)]$

$&rArr;$  $a = b$ or ${c^2}a + {c^2}b - {a^2}b - {a^2}c = 0$

$&rArr;$  ${c^2}a + {c^2}b - {a^2}b - {a^2}c = 0$

$&rArr;$  $ac\,(c - a) = b\,({a^2} - {c^2})$

$&rArr;$  $ac = - b\,(c + a)$

$&rArr;$  $ - ac = b.2b$

$&rArr;$  ${b^2} = ( - a/2)\,c$,

$	herefore - a/2,b,c$ are in $G.P.$

If $a,\,b,\,c$ are in $A.P.$ and ${a^2},\,{b^2},{c^2}$ are in $H.P.$, then

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