Animated Solution for Mathematics - Sequence and Series: Let $a_1,\frac{a_2}{2},\frac{a_3}{2^2},\dots ,\frac{a_{10}}{2^9}$ be a G.P. of common ratio $\frac{1}{\sqrt{2}}$. If $a_1+a_2+\cdots +a_{10}=62$, then $a_1$ is equal to :

Unmasking the Hidden Sequence

My dear student, welcome to the world of sequences and series. In the JEE Advanced arena, problems are rarely presented in their simplest form. They are often disguised, wearing masks of complexity to test your ability to see through the noise.

Today, we are going to peel back the layers of a seemingly intimidating problem and reveal the elegant structure hiding underneath.

We are given a sequence: $a_1,\frac{a_2}{2},\frac{a_3}{2^2},\dots ,\frac{a_{10}}{2^9}$. We are told this is a Geometric Progression (G.P.) with a common ratio of $r=\frac{1}{\sqrt{2}}$.

At first glance, your brain might want to panic. You see fractions, you see powers of two, and you see a sequence that doesn't look like the standard $a,ar,a{r}^2$ format. But take a deep breath. Let us define a new sequence, $b_n$, where $b_n=\frac{a_n}{2^{n-1}}$.

The Algebraic Bridge

By defining $b_n=\frac{a_n}{2^{n-1}}$, we have effectively stripped away the disguise. We know that for any G.P., the $n$-th term is given by $b_n=b_1\cdot r^{n-1}$.

Since $b_1=a_1$ and $r=\frac{1}{\sqrt{2}}$, we can write:

This equation is our bridge. It connects the world of the given sequence to the world of the sequence we actually care about: $a_n$. Our goal is to isolate $a_n$. To do this, we multiply both sides by $2^{n-1}$.

Look at the symmetry here! Both terms on the right side are raised to the power of $n-1$. In mathematics, whenever you see exponents matching, your intuition should scream, "Combine them!" We can group the bases together:

The Beauty of Simplification

Now, let us look at that fraction inside the parenthesis: $\frac{2}{\sqrt{2}}$. Many students freeze here, but remember your basic surd properties. We know that $2=\sqrt{2}\cdot \sqrt{2}$.

Therefore, $\frac{\sqrt{2}\cdot \sqrt{2}}{\sqrt{2}}$ simplifies beautifully to just $\sqrt{2}$. Suddenly, the complexity collapses. Our general term becomes:

Stop for a moment and appreciate what we have just discovered. This is the definition of a G.P. with a first term $a_1$ and a common ratio $R=\sqrt{2}$.

The sequence $a_1,a_2,\dots ,a_{10}$ is not just a random collection of numbers; it is a perfectly ordered G.P. The problem has transformed from a confusing mess into a standard summation problem.

The Final Calculation

We are given that the sum of these ten terms is $62$. That is, $S_{10}=62$. The formula for the sum of the first $n$ terms of a G.P. is:

Substituting our values ($n=10$, $R=\sqrt{2}$, $S_{10}=62$), we get:

Now, let us tackle the exponent. We need to evaluate $(\sqrt{2}{)}^{10}$. As we discussed, $\sqrt{2}=2^{1/2}$. So, $(\sqrt{2}{)}^{10}=(2^{1/2}{)}^{10}=2^5=32$.

Substituting this back into our equation:

This is the moment of truth. We divide both sides by $31$. Since $62=2\cdot 31$, the left side becomes $2$:

Finally, we isolate $a_1$ by multiplying both sides by $(\sqrt{2}-1)$:

The Mindset of a Master

And there you have it. We started with a sequence that looked like a tangled knot, and through systematic, logical steps, we unraveled it to find a clean, elegant solution.

This is the essence of JEE Advanced physics and mathematics. It is not about memorizing formulas; it is about recognizing patterns, simplifying expressions, and having the patience to follow the logic to its conclusion. You have the tools, you have the intuition—now go forth and solve the next one with this same confidence!