Question

If the sum of the first n terms of the series $\sqrt{3}+\sqrt{75}+\sqrt{243}+\sqrt{507}+...$ is $435\sqrt{3}$, then n equals
(a) 18
(b) 13
(c) 29
(d) 15

Answer 1

Hint: First of all, we need to find out which kind of series is this. We will do this by simplifying the terms given in the roots and taking the numbers out of the root whenever possible. This will be done by factorising the number given as square roots. Once we get the series and can decide which progression is it, then we can apply the formula for that progression. Two most common progressions are arithmetic progression (AP) and geometric progression (GP). The formula for sum of first n terms of an arithmetic progression is given by the relation ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$, where a is the first term and d is the common difference, whereas the sum of n terms of geometric progression ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$, where a is the first term and r is the common ratio of the series.

Complete step by step answer:
The series given to us is $\sqrt{3}+\sqrt{75}+\sqrt{243}+\sqrt{507}+...$
The second term is $\sqrt{75}$. 75 can be factored as 5*5*3.
Thus $\sqrt{75}=5\sqrt{3}$
The third term of the series is $\sqrt{243}$. 243 can be factored as 9*9*3.
Thus $\sqrt{243}=9\sqrt{3}$
The third term of the series is $\sqrt{507}$. 243 can be factored as 13*13*3.
Thus $\sqrt{507}=13\sqrt{3}$
Therefore, the series can be written as $\sqrt{3}+5\sqrt{3}+9\sqrt{3}+13\sqrt{3}+...$
By observation, we can say that the series is in Arithmetic Progression.
We can safely assume that further terms will also be some multiple of $\sqrt{3}$.
Thus, we can take $\sqrt{3}$ common from all terms.
$\Rightarrow \sqrt{3}\left( 1+5+9+13+... \right)=435\sqrt{3}$
$\sqrt{3}$ gets divided out from both sides.
$\Rightarrow 1+5+9+13+...=435$
Therefore, we get a new AP with the first term as 1 and common difference as 4.
As we know, the formula for sum of n terms of AP is given as ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$.
Substitute ${{S}_{n}}=435$, $a=1$, $d=4$ and find the value of n.
$\begin{align}
  & \Rightarrow 435=\dfrac{n}{2}\left[ 2\left( 1 \right)+\left( n-1 \right)\left( 4 \right) \right] \\
 & \Rightarrow 870=n\left[ 2+4n-4 \right] \\
 & \Rightarrow 4{{n}^{2}}-2n-870=0 \\
 & \Rightarrow 2{{n}^{2}}-n-435=0 \\
\end{align}$
We will solve this quadratic equation by factorisation method.
To solve a quadratic equation, we need to split the middle term such that the product of coefficients of the terms after splitting must be equal to the product of the coefficients of the second order variable and constant and their sum must be equal to coefficient of first order variable. Thus, the middle term $-n$ is written as $-30n+29n$.
$\begin{align}
  & 2{{n}^{2}}-30n+29n-435=0 \\
 & 2n\left( n-15 \right)+29\left( n-15 \right)=0 \\
 & \left( n-15 \right)\left( 2n+29 \right)=0 \\
\end{align}$
Thus n = 15 or $n=\dfrac{-29}{2}$, but we know that n is a positive integer. Thus n = 15.

So, the correct answer is “Option D”.

Note: Students can verify the options once they get a quadratic equation. The option which satisfies the equation will be the correct option. This may save them the steps and time required to solve a quadratic equation.