Question

Find 15th term of sequence √3,1/√3,1/3√3​

Answer 1

Solution :-

We have got the sequence :-

$\sqrt{3} ,\dfrac{1}{\sqrt{3}} , \dfrac{1}{3\sqrt{3}} , ...$

Now as we are able to see it's a G.P ( Geometric Progression)

We know nth term in a G.P

$= a(r)^{n-1}$

Where

a = First term

r = common ratio

Now Common ratio

= 2nd term by 1st term

$= \dfrac{\dfrac{1}{\sqrt{3}}}{\sqrt{3}}$

$= \dfrac{1}{3}$

So we have got our common ratio and first term.

Now putting it into the equation for nth term.

$= \sqrt{3}. \left(\dfrac{1}{3}\right)^{15-1}$

$= \sqrt{3}. \left(\dfrac{1}{3}\right)^{14}$

$= \sqrt{3}.\dfrac{1}{3^{14}}$

$= \dfrac{1}{\sqrt{3}.3^{13}}$

So Answer

$\huge{\boxed{\boxed{\sf{ = \dfrac{1}{\sqrt{3}.3^{13}}}}}}$

Answer 2

Solution :-

We have got the sequence :-

$\sqrt{3} ,\dfrac{1}{\sqrt{3}} , \dfrac{1}{3\sqrt{3}}$ , ...

Now as we are able to see it's a G.P ( Geometric Progression)

We know nth term in a G.P

= $a(r)^{n-1}$

Where

a = First term

r = common ratio

Now Common ratio

= 2nd term by 1st term

= $\dfrac{\dfrac{1}{\sqrt{3}}}{\sqrt{3}}$

= $\dfrac{1}{3}$

So we have got our common ratio and first term.

Now putting it into the equation for nth term.

= $\sqrt{3}. \left(\dfrac{1}{3}\right)^{15-1}$

= $\sqrt{3}. \left(\dfrac{1}{3}\right)^{14}$

= $\sqrt{3}.\dfrac{1}{3^{14}}$

= $\dfrac{1}{\sqrt{3}.3^{13}}$

So Answer

$\huge{\boxed{\sf{ = \dfrac{1}{\sqrt{3}.3^{13}}}}}$