Question

A certain number is inserted between 3 and P so that the three numbers form of an A.P.is middle term is diminished by 6 then the number are in G.P. then P=

• A. 3
• B. 15
• C. 18
• D. 27

Answer 1

Answer: (A), (D)

The correct options are
A 3
D 27

If a number is listed between $3$ & $P$

$3,a,p⟶AP$

$3+p=2a$

$a=\frac{3+p}{2}$

Now if $3,a-6,p$ then $⟶GP$

$\frac{a-6}{3}=\frac{p}{a-6}{(a-6)}^2=3P{a}^2+36-12a=3P{\left(\frac{3+P}{2}\right)}^2+36-12\left(\frac{3+P}{2}\right)=3P$

$P^2-30P+81=0$

$P^2-27P-3P+81=0$

$(P-3)(P-27)=0$

$\therefore P=3,27.$