Question

find the 10 th term of ap √3, √12,√27,...​

Answer 1

Answer:

√300 = 10√3

Step-by-step explanation:

Note :

Tn = a + ( n - 1 ) d

a = 1st term = √3

d = common difference = T2 - T1 = √12 - √3

= √( 4*3 ) - √3 = 2√3 - √3 = 1√3 = √3

n = nth term = 10th term

Tn = a + ( n - 1 ) d

=> T10 = √3 + ( 10 - 1 ) √3

=> T10 = √3 + 9√3

=> T10 = 10√3 = √(100*3) = √300

Answer 2

Answer:

$\displaystyle{\boxed{\red{\sf\:t_{10}\:=\:10\:\sqrt{3}}}}$

Step-by-step-explanation:

We have given an arithmetic progression.

We have to find the 10ᵗʰ term of the given AP.

The given AP is $\displaystyle{\sf\:\sqrt{3}\:,\:\sqrt{12}\:,\sqrt{27}\:,\:\dots}$

Now,

$\displaystyle{\bullet\:\sf\:a\:=\:t_1\:=\:\sqrt{3}}$

$\displaystyle{\bullet\:\sf\:t_2\:=\:\sqrt{12}}$

$\displaystyle{\bullet\:\sf\:t_3\:=\:\sqrt{27}}$

Now, we know that,

$\displaystyle{\sf\:Common\:difference\:(\:d\:)\:=\:t_2\:-\:t_1}$

$\displaystyle{\implies\sf\:d\:=\:\sqrt{12}\:-\:\sqrt{3}}$

$\displaystyle{\implies\sf\:d\:=\:\sqrt{4\:\times\:3}\:-\:\sqrt{3}}$

$\displaystyle{\implies\sf\:d\:=\:\sqrt{2\:\times\:2\:\times\:3}\:-\:\sqrt{3}}$

$\displaystyle{\implies\sf\:d\:=\:2\:\sqrt{3}\:-\:\sqrt{3}}$

$\displaystyle{\implies\sf\:d\:=\:(\:2\:-\:1\:)\:\sqrt{3}}$

$\displaystyle{\implies\boxed{\red{\sf\:d\:=\:\sqrt{3}}}}$

Now, we know that,

$\displaystyle{\pink{\sf\:t_n\:=\:a\:+\:(\:n\:-\:1\:)\:\times\:d}\sf\:\:\:-\:-\:-\:[\:Formula\:]}$

$\displaystyle{\implies\sf\:t_{10}\:=\:\sqrt{3}\:+\:(\:10\:-\:1\:)\:\times\:\sqrt{3}}$

$\displaystyle{\implies\sf\:t_{10}\:=\:\sqrt{3}\:+\:9\:\times\:\sqrt{3}}$

$\displaystyle{\implies\sf\:t_{10}\:=\:\sqrt{3}\:+\:9\:\sqrt{3}}$

$\displaystyle{\implies\sf\:t_{10}\:=\:(\:1\:+\:9\:)\:\sqrt{3}}$

$\displaystyle{\implies\underline{\boxed{\red{\sf\:t_{10}\:=\:10\:\sqrt{3}}}}}$