Question

If $a,b,c$ are in GP and $a,p,q$ are in AP such that $2a,b+p,c+q$ are in GP then the common difference of AP is

• A. $\sqrt{2a}$
• B. $(\sqrt{2}+1)(a-b)$
• C. $\sqrt{2}(a+b)$
• D. $(\sqrt{2}-1)(b-a)$

Answer 1

Answer: (B), (D)

The correct options are
B $(\sqrt{2}+1)(a-b)$
D $(\sqrt{2}-1)(b-a)$

Let common difference of the A.P is $d$.
Given:
$b^2=ac$ .......(1)
$2p=a+q$ .......(2)
$(b+p{)}^2=(c+q)2a$ .......(3)
(3)-(1) we get, $\Rightarrow p(2b+p)=ac+2aq$
$\Rightarrow b^2-\left(2p\right)b+2aq-p^2=0$ from (1)
$\Rightarrow b^2-(a+q)+\frac{8aq-(a+q{)}^2}{4}=0$
$\therefore b=\frac{a+q\pm \sqrt{a^2+q^2+2aq+a^2+q^2+2aq-8aq}}{2}$
$\Rightarrow 2b=a+q\pm \sqrt{2a^2+2q^2-4aq}$
$=a+q\pm \sqrt{2}(q-a)$
$=a(1\mp \sqrt{2})+q(1\pm \sqrt{2})$
$=a(1\mp \sqrt{2})+(a+2d)(1\pm \sqrt{2})$
$\Rightarrow 2b=2a+2d(1\pm \sqrt{2})$
$\Rightarrow d=\frac{b-a}{1\pm \sqrt{2}}$
or $d=(\sqrt{2}-1)(b-a),(\sqrt{2}+1)(a-b)$
Hence, Options B and D are correct.