Given:
Sum of GP = \( 39 + 13\sqrt{3}\)
Where, a =\( \sqrt{3}\), r = \( 3\sqrt{3}\) = \( \sqrt{3}\), n = ?
By using the formula,
Sum of GP for n terms =
\( 39 + 13\sqrt{3} = \dfrac{\sqrt{3} (\sqrt{3}n – 1)}{(\sqrt{3} – 1)}\)
Let us simplify we get,
\( 39\sqrt{3} – 39 + 13(3) – 13\sqrt{3} = \sqrt{3} (\sqrt{3}n – 1)\)
\( 27\sqrt{3} = \sqrt{3}n+1\)
\( \sqrt{3}^6 \sqrt{3} = \sqrt{3}n+1\)
6+1 = n + 1
7 = n + 1
7 – 1 = n
6 = n
6 terms are required to make a sum of \( 39 + 13\sqrt{3}\)
Answered by Sakshi | 1 year agoConstruct a quadratic in x such that A.M. of its roots is A and G.M. is G.
Find the geometric means of the following pairs of numbers:
(i) 2 and 8
(ii) a3b and ab3
(iii) –8 and –2
Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).