Question

Three numbers form an increasing G. P. If the middle term is doubled , then the new numbers are in A. P. Find the common ratio of the G. P.

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{\boxed { \huge \rm \red{ Solution}}}}$

Let numbers in G.P. be a, ar and ar²

$\mathrm {2(2ar)=a+ar^{2} \ \ (Since, \ a \neq 0 )}$

⇒ 4r = 1 + r²  

⇒ r²  − 4r + 1 = 0

Now,

$\sf {x= \frac{-b \± \sqrt{b^{2} -4ac}}{2a} }\\\\\\\sf {\rightarrowtail \ r=2 \± \sqrt{3}}\\\\\\\sf {\rightarrowtail \ r=2 + \sqrt{3}}$

$\sf {\because \ 2 - \sqrt{3} < 1}\\\\\\\sf {\therefore \ For \ r=2 - \sqrt{3}, \ the \ G.P. \ will \ not \ be \ increasing}$

Answer 2

Step-by-step explanation:

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{\boxed { \huge \rm \red{ Solution}}}}$

Let numbers in G.P. be a, ar and ar²

$\mathrm {2(2ar)=a+ar^{2} \ \ (Since, \ a \neq 0 )}$

⇒ 4r = 1 + r²  

⇒ r²  − 4r + 1 = 0

Now,

$\sf {x= \frac{-b \± \sqrt{b^{2} -4ac}}{2a} }\\\\\\\sf {\rightarrowtail \ r=2 \± \sqrt{3}}\\\\\\\sf {\rightarrowtail \ r=2 + \sqrt{3}}$

$\sf {\because \ 2 - \sqrt{3} < 1}\\\\\\\sf {\therefore \ For \ r=2 - \sqrt{3}, \ the \ G.P. \ will \ not \ be \ increasing}$