Question

if sum of 1st 10 terms of an AP
$\sqrt{3} \sqrt{12} \sqrt{27} \sqrt{48} ......is \: m \sqrt{3}$
then m is ......​

Answer 1

Answer:

30

Step-by-step explanation:

the A. P can be written as√3, 2√3, 3√3, 4√3 so the common difference is √3, A is √3, sum of 10 terms is

5(2√3+(4√3))

5(6√3)

30√3

so m is 30

Answer 2

$Given:-\\n = 10\\AP = \sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48} ,..........m\sqrt{3} \\s_{n} = m\sqrt{3}$

$AP = a + ( n - 1) d\\ = \sqrt{3} + 9(\sqrt{12} - \sqrt{3} )\\ = 10\sqrt{3}$

$therefore \\s_n = (\frac{a + a_n}{2} n\\ = 55\sqrt{3} \\\\therfore m = 55$