Let a_1, fraca_22, fraca_32^2, ldots, fraca_102^9$a_1, \frac{a_2}{2}, \frac{a_3}{2^2}, \ldots, \frac{a_{10}}{2^9}$ be a G.P. of common ratio \frac{1}{\sqrt{2}}. If a_1 + a_2 + ldots + a_10 = 62$a_1 + a_2 + \ldots + a_{10} = 62$, then a_1$a_1$ is equal to:

### Related Formula textSum of G.P. S_n = fraca(r^n - 1)r - 1 text (for r > 1)$$\text{Sum of G.P. } S_n = \frac{a(r^n - 1)}{r - 1} \text{ (for } r > 1)$$ ### Core Logic The given sequence is a G.P. with ratio frac1sqrt2$\frac{1}{\sqrt{2}}$. fraca_2/2a_1 = frac1sqrt2 implies a_2 = a_1 sqrt2$$\frac{a_2/2}{a_1} = \frac{1}{\sqrt{2}} \implies a_2 = a_1 \sqrt{2}$$ fraca_3/2^2a_2/2 = frac1sqrt2 implies fraca_32 a_2 = frac1sqrt2 implies a_3 = a_2 sqrt2 = a_1 (sqrt2)^2$$\frac{a_3/2^2}{a_2/2} = \frac{1}{\sqrt{2}} \implies \frac{a_3}{2 a_2} = \frac{1}{\sqrt{2}} \implies a_3 = a_2 \sqrt{2} = a_1 (\sqrt{2})^2$$ Thus, a_1, a_2, a_3, dots, a_10$a_1, a_2, a_3, \dots, a_{10}$ forms a standard G.P. with first term a_1$a_1$ and common ratio R = sqrt2$R = \sqrt{2}$. ### Step 1: Calculate the Sum Sum of this new sequence is S_10 = 62$S_{10} = 62$. S_10 = fraca_1 left( (sqrt2)^10 - 1 right)sqrt2 - 1 = 62$$S_{10} = \frac{a_1 \left( (\sqrt{2})^{10} - 1 \right)}{\sqrt{2} - 1} = 62$$ Since (sqrt2)^10 = 2^5 = 32$(\sqrt{2})^{10} = 2^5 = 32$: 62 = fraca_1 (32 - 1)sqrt2 - 1$$62 = \frac{a_1 (32 - 1)}{\sqrt{2} - 1}$$ 62 = frac31 a_1sqrt2 - 1$$62 = \frac{31 a_1}{\sqrt{2} - 1}$$ 2 = fraca_1sqrt2 - 1$$2 = \frac{a_1}{\sqrt{2} - 1}$$ a_1 = 2(sqrt2 - 1)$$a_1 = 2(\sqrt{2} - 1)$$ ### Pattern Recognition If a sequence b_n = fraca_nk^n-1$b_n = \frac{a_n}{k^{n-1}}$ is a G.P. with ratio r$r$, then the base sequence a_n$a_n$ is inherently a G.P. with ratio R = kr$R = kr$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequence and Series