How many terms of the sequence $$\sqrt{3}$$, 3, $$3 \sqrt{3}$$,… must be taken to make the sum 39+ $$13 \sqrt{3}$$ ?

Asked by Sakshi | 1 year ago |  71

##### Solution :-

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 ago

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