Question

a, b, c $\in R$ and a, b, c in A.P. match the entries of column-I with those or column-II

Answer 1

(a) Given $2b=a+c,2b^2=a^2+c^2$
$2b^2=(a+c{)}^2-2ac=4b^2-2ac$
$\therefore 2ac=4b^2-2b^2=2b^2$ $\therefore b^{2}ac$
$2b^2=a^2+c^2$ or $2ac=a^2+c^2$ or $(a-b{)}^2=0$
Hence $a=b=c$ $\therefore \left(a\right)\to \left(p\right)$
(b) $a^2,b^2,c^2$ are in G.P. $⟹(b^2{)}^2=a^2{c}^2$
$b^2=ac,b^2=-ac$
$b^2=ac⟹a,b,c$ are in G.P.
But $a,b,c$are in A.P. as well $\therefore a=b=c$
$\therefore \left(b\right)\to \left(p\right)$
(c) $b^2=\frac{2a^2{c}^2}{a^2+c^2}$ or $b^2\left[\right(a+b{)}^2-2ac]=2a^2{c}^2$
$\left(\frac{a+c}{2}\right)\left[\right(a+b{)}^2-2ac]=2a^2{c}^2$
or $(a+c{)}^4-2ac(a+c^2)-8a^{2}C62=0$
or $\left[\right(a+c{)}^2-4ac\left]\right[(a+c{)}^2-2ac]=0$
or $(a-c{)}^2=0$ or $(a+c{)}^2=-2ac$
or $(a-c)$ or $(2b{)}^2=-2ac$
or $a=b=c$ $\because 2b=a+c$ and $b^2=-\frac{a}{2}c$
$\therefore a,b,c$ are in A.P. or $-\frac{a}{2},b,c$
or $a,b,-\frac{c}{2}$ are in G.P. $\therefore \left(c\right)\to (p,q,r)$.
(d) $a+b+c=\frac{3}{2}⟹b=\frac{1}{2}$
$\therefore \left(d\right)\to \left(s\right)$