Three numbers are in an increasing geometric progression with common ratio r. If the middle number is doubled, then the new numbers are in an…

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​