JEE Main 2021 — Mathematics Algebra

Three numbers are in an increasing geometric progression with common ratio $r.$r.

Let first term is $a$a

$$\Rightarrow \frac{a}{r},a,ar\in G.P$$⇒ra​,a,ar∈G.P

Given that if middle term is doubled

$$\Rightarrow \frac{a}{r},2a,ar\in A.P$$⇒ra​,2a,ar∈A.P

If $a,b,c\in A.P\Rightarrow 2b=a+c$a,b,c∈A.P⇒2b=a+c

$$4a=ar+\frac{a}{r}$$4a=ar+ra​

$$4=r+\frac{1}{r}$$4=r+r1​

$${r}^{2}+1=4r$$r2+1=4r

$${r}^{2}-4r+1=0$$r2−4r+1=0

If$a{x}^{2}+bx+c=0\Rightarrow x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2ac}$ax2+bx+c=0⇒x=2ac−b±b2−4ac​​

$$\Rightarrow r=\frac{4\pm \sqrt{12}}{2}=2+\sqrt{3},2-\sqrt{3}$$⇒r=24±12​​=2+3​,2−3​

But it is an increasing $G.P$G.P

$$\therefore r=2+\sqrt{3}$$∴r=2+3​

Given that Fourth term of $G.P\Rightarrow {t}_{4}=a{r}^{2}=3{r}^{2}$G.P⇒t4​=ar2=3r2

$$\Rightarrow a=3.$$⇒a=3.

Common difference of an $A.P=d={t}_{2}-{t}_{1}$A.P=d=t2​−t1​

$$d=2a-\frac{a}{r}=a(2-\frac{1}{r})=3(2-\frac{1}{2+\sqrt{3}})$$d=2a−ra​=a(2−r1​)=3(2−2+3​1​)

$$=3(2-2+\sqrt{3})=3\sqrt{3}$$=3(2−2+3​)=33​

Hence ${r}^{2}-d=(2+\sqrt{3}{)}^{2}-(3\sqrt{3})$r2−d=(2+3​)2−(33​)

$$=4+3+4\sqrt{3}-3\sqrt{3}$$=4+3+43​−33​

$$=7+\sqrt{3}$$=7+3​