Question

Show that root 13, root 12 root 27, root 48 is in AP

Answer 1

it's first of all

$\sqrt{3}$

not

$\sqrt{13}$

we will obtain an sequence

$\sqrt{3}$

,

$2 \sqrt{3}$

,

$3 \sqrt{3}$

,

$4 \sqrt{3}$

so a1=

$\sqrt{3}$

a2=

$2 \sqrt{3}$

a3=

$3 \sqrt{3}$

and a4=

$4 \sqrt{3}$

so let d be the common difference

now

a2-a1=

$2 \sqrt{3 } - \sqrt{3 } = \sqrt{3}$

similarly

a3-a2=

$3 \sqrt{3} - 2 \sqrt{3} = \sqrt{3}$

and a4-a3=

$4 \sqrt{3} - 3 \sqrt{3} = \sqrt{3}$

therefore

common difference is

$\sqrt{3}$

so

the sequence is ap