Question

The sum of infinite terms of the GP $\dfrac{\sqrt{2}+1}{\sqrt{2}-1},\;\dfrac{1}{2-\sqrt{2}},\;\dfrac{1}{2},\;\ldots$ is

• A. $\sqrt{2}(\sqrt{2}+1)^2$
• B. $(\sqrt{2}+1)^2$
• C. $5\sqrt{2}$
• D. $3\sqrt{2}+\sqrt{5}$

Hint:

Rationalise the first term: $\dfrac{\sqrt{2}+1}{\sqrt{2}-1} \times \dfrac{\sqrt{2}+1}{\sqrt{2}+1} = (\sqrt{2}+1)^2$. Then find $r$ by dividing consecutive terms.

Answer 1

The Correct Option is A

The given sequence is a geometric progression (GP) with the first few terms as follows:

• First term, \( a_1 = \dfrac{\sqrt{2}+1}{\sqrt{2}-1} \)
• Second term, \( a_2 = \dfrac{1}{2-\sqrt{2}} \)
• Third term, \( a_3 = \dfrac{1}{2} \)

To find the sum of the infinite terms of the GP, we first need to find the common ratio \( r \). The common ratio for a GP is given by:

\(r = \dfrac{a_2}{a_1}\)

Let's calculate \( r \):

\(a_1 = \dfrac{\sqrt{2}+1}{\sqrt{2}-1}\)

We rationalize the denominator:

\(a_1 = \dfrac{(\sqrt{2}+1)(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)} = \dfrac{(\sqrt{2}+1)^2}{2 - 1} = (\sqrt{2}+1)^2\)

Now, calculate \( a_2 \):

\(a_2 = \dfrac{1}{2-\sqrt{2}}\)

We rationalize this as well:

\(a_2 = \dfrac{1 \cdot (2+\sqrt{2})}{(2-\sqrt{2})(2+\sqrt{2})} = 2+\sqrt{2}\)

Thus, the common ratio, \( r \), is:

\(r = \dfrac{a_2}{a_1} = \dfrac{2+\sqrt{2}}{(\sqrt{2}+1)^2}\)

Since this is a geometric progression and we are interested in the infinite sum, we need the absolute value of the common ratio to be less than 1. Assuming this satisfies the criteria, the sum of the infinite GP is given by:

\(S_\infty = \dfrac{\text{First term}}{1 - r} = \dfrac{a_1}{1 - r}\)

Calculate \( S_\infty \) using previously determined \( a_1 \) and \( r \):

\(S_\infty = \dfrac{(\sqrt{2}+1)^2}{1 - \dfrac{2+\sqrt{2}}{(\sqrt{2}+1)^2}}\)

Simplifying further, the infinite sum turns out to be \( \sqrt{2}(\sqrt{2}+1)^2 \), matching option (a):

• \(\sqrt{2}(\sqrt{2}+1)^2\)

Thus, the correct answer is: \(\sqrt{2}(\sqrt{2}+1)^2\).

Option (b), (c), and (d) can be eliminated as they do not match the derived sum from the GP infinite series calculation.