If a, b, c are in A.P., $${a^2}$$, $${b^2}$$, $${c^2}$$ are in H.P., then prove that either a = b = c or a, b, $${ - {c \over 2}}$$ form a G.P.

Prove that
$${2^k}\left( {\matrix{ n \cr 0 \cr } } \right)\left( {\matrix{ n \cr k \cr } } \right) - {2^{^{k - 1}\left( {\matrix{ n \cr 2 \cr } } \right)}}\left( {\matrix{ n \cr 1 \cr } } \right)\left( {\matrix{ {n - 1} \cr {k - 1} \cr } } \right)$$
$$ + {2^{k - 2}}\left( {\matrix{ {n - 2} \cr {k - 2} \cr } } \right) - .....{\left( { - 1} \right)^k}\left( {\matrix{ n \cr k \cr } } \right)\left( {\matrix{ {n - k} \cr 0 \cr } } \right) = {\left( {\matrix{ n \cr k \cr } } \right)^ \cdot }$$

If $${x^2} + \left( {a - b} \right)x + \left( {1 - a - b} \right) = 0$$ where $$a,\,b\, \in \,R$$ then find the values of a for which equation has unequal real roots for all values of $$b$$.

If $${z_1}$$ and $${z_2}$$ are two complex numbers such that $$\,\left| {{z_1}} \right| < 1 < \left| {{z_2}} \right|\,$$ then prove that $$\,\left| {{{1 - {z_1}\overline {{z_2}} } \over {{z_1} - {z_2}}}} \right| < 1$$.