$$\begin{gathered} \text{Let}\mathrm{ a}\mathrm{1,}\frac{a_2}{2}\frac{a_3}{2}\text{…}\frac{a_{10}}{2}\text{be a G.P of common ratio}\frac{1}{\sqrt{2}}.\text{If}{\mathrm{ a}}_1+a_2+\text{…}+a_{10}=62\text{, then}{a}_1\text{is equal to} \\ \end{gathered}$$

The correct answer is 2(\sqrt{2}-1).

Step 1: Understand the given Geometric Progression (G.P.)
Let the terms of the given geometric progression be represented by b_1,b_2,b_3,\dots ,b_{10}, where each term is given by:

b_n=\frac{a_n}{2^{n-1}}

The common ratio of this G.P. is given as r=\frac{1}{\sqrt{2}}. Therefore, the general term b_n can be expressed in terms of the first term b_1=a_1 as:

b_{\mathrm{n}}=a_1·{\left(\frac{1}{\sqrt{2}}\right)}^{n-1}

Step 2: Relate a_n to a_1
Equating the two expressions for b_n:

\frac{a_n}{2^{n-1}}

Multiplying both sides by 2^{n-1}:

a_n=a_1·{\left(2·\frac{1}{\sqrt{2}}\right)}^{n-1}=a_1{\left(\sqrt{2}\right)}^{n-1}

This shows that the sequence a_1,a_2,\dots ,a_{10} is a new G.P. with first term a_1 and common ratio R=\sqrt{2}.

Step 3: Calculate the sum of the terms
The formula for the sum of the first 10 terms of a G.P. is:

S_{10}=a_1+a_2+\cdots +a_{10}=a_1\frac{R^{10}-1}{R-1}

Substitute R=\sqrt{2} into the equation:

{\left(\sqrt{2}\right)}^{10}=2^5=32

Thus, the sum simplifies to:

S_{10}=a_1\frac{32-1}{\sqrt{2}-1}=\frac{31a_1}{\sqrt{2}-1}

Step 4: Solve for a_1
We are given that S_{10}=62:

\frac{31a_1}{\sqrt{2}-1}=62

Divide both sides by 31:

\frac{a_1}{\sqrt{2}-1}=2

Multiplying by (\sqrt{2}-1) gives:

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