If the sum of the first n terms of the series | vedclass

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(B) The given series is $\sqrt{3} + \sqrt{75} + \sqrt{243} + \sqrt{507} + \dots$This can be written as $\sqrt{3} + 5\sqrt{3} + 9\sqrt{3} + 13\sqrt{3}

Vedclass · Accepted Answer

$15$

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(B) The given series is $\sqrt{3} + \sqrt{75} + \sqrt{243} + \sqrt{507} + \dots$This can be written as $\sqrt{3} + 5\sqrt{3} + 9\sqrt{3} + 13\sqrt{3} + \dots$This is an Arithmetic Progression with first term $a = \sqrt{3}$ and common difference $d = 4\sqrt{3}$.The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.Given $S_n = 435\sqrt{3}$,we have:$\frac{n}{2}[2\sqrt{3} + (n-1)4\sqrt{3}] = 435\sqrt{3}$Divide both sides by $\sqrt{3}$:$\frac{n}{2}[2 + 4n - 4] = 435$$n(2n - 2) = 870$$2n^2 - 2n - 870 = 0$$n^2 - n - 435 = 0$Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$n = \frac{1 \pm \sqrt{1 - 4(1)(-435)}}{2} = \frac{1 \pm \sqrt{1 + 1740}}{2} = \frac{1 \pm \sqrt{1741}}{2}$Wait,checking the series calculation: $1, 5, 9, 13$ are terms of an $AP$ with $a=1, d=4$. Sum is $\frac{n}{2}[2(1) + (n-1)4] = \frac{n}{2}[4n - 2] = n(2n-1) = 435$.$2n^2 - n - 435 = 0$.$n = \frac{1 \pm \sqrt{1 + 4(2)(435)}}{4} = \frac{1 \pm \sqrt{1 + 3480}}{4} = \frac{1 \pm \sqrt{3481}}{4} = \frac{1 \pm 59}{4}$.Since $n$ must be positive,$n = \frac{60}{4} = 15$.

If the sum of the first $n$ terms of the series $\sqrt{3} + \sqrt{75} + \sqrt{243} + \sqrt{507} + \dots$ is $435\sqrt{3}$,then $n$ equals:

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