Let a₁, a₂/2, a₃/2², …, a₁₀/2⁹ be a G.P. of common ratio 1/√2. If a₁ + a₂ + … + a₁₀ = 62, then a₁ is equal to

A Geometric Progression (GP) is a sequence where each term is obtained by multiplying the previous term by a fixed non-zero number called the common ratio $r$.

General form: $a,\ ar,\ ar^2,\ ar^3,\ \ldots$

$n$-th term: $T_n = a\cdot r^{n-1}$

Sum of $n$ terms ($r \neq 1$): $$S_n = \frac{a(r^n – 1)}{r-1} \quad \text{if } r > 1$$ $$S_n = \frac{a(1-r^n)}{1-r} \quad \text{if } r < 1$$ Both forms give the same result — choose whichever avoids negative denominators for cleaner arithmetic.

This problem presents a modified GP: instead of the original $a_n$, the terms $\dfrac{a_n}{2^{n-1}}$ form a GP. This is a common JEE trick.

Key logic: If $b_n = \dfrac{a_n}{f(n)}$ is a GP with ratio $r$, then: $$\frac{b_{n+1}}{b_n} = r \Rightarrow \frac{a_{n+1}/f(n+1)}{a_n/f(n)} = r \Rightarrow \frac{a_{n+1}}{a_n} = r \cdot \frac{f(n+1)}{f(n)}$$ Here $f(n) = 2^{n-1}$, so $f(n+1)/f(n) = 2$, giving $a_{n+1}/a_n = (1/\sqrt{2})\cdot 2 = \sqrt{2}$.

So the original $\{a_n\}$ is a GP with ratio $\sqrt{2}$ — completely different from the given ratio $1/\sqrt{2}$.

Since $\sqrt{2} = 2^{1/2}$, powers of $\sqrt{2}$ are powers of 2 divided by appropriate roots:

$(\sqrt{2})^1 = \sqrt{2}$ $(\sqrt{2})^2 = 2$ $(\sqrt{2})^4 = 4$ $(\sqrt{2})^6 = 8$ $(\sqrt{2})^8 = 16$ $(\sqrt{2})^{10} = 32$ $(\sqrt{2})^{2k} = 2^k$

The key result here: $(\sqrt{2})^{10} = 2^5 = 32$. This makes the numerator $r^n – 1 = 32 – 1 = 31$, which divides cleanly into 62, giving a clean answer.

The equation $\dfrac{31a_1}{\sqrt{2}-1} = 62$ gives $a_1 = \dfrac{2(\sqrt{2}-1)}{1}$ directly — no rationalisation needed since the answer itself contains $\sqrt{2}-1$.

However, if you need to verify: multiply numerator and denominator of $\dfrac{2}{\sqrt{2}-1}$ by $(\sqrt{2}+1)$: $$\frac{2(\sqrt{2}+1)}{(\sqrt{2})^2-1^2} = \frac{2(\sqrt{2}+1)}{1} = 2\sqrt{2}+2$$ This would be $a_1$ if we rationalised, but the answer $2(\sqrt{2}-1)$ is the correct form.

Quick check: $2(\sqrt{2}-1) \approx 2(1.414-1) = 2(0.414) \approx 0.828 > 0$ ✓ (positive first term makes sense for a positive GP)

With $a_1 = 2(\sqrt{2}-1)$ and ratio $r = \sqrt{2}$: $$S_{10} = \frac{2(\sqrt{2}-1) \cdot (32-1)}{\sqrt{2}-1} = 2 \times 31 = 62 \ ✓$$ The $(\sqrt{2}-1)$ factors cancel beautifully, confirming the answer. This type of cancellation is a hallmark of well-constructed JEE problems.

1. Product of terms equidistant from ends: In a finite GP, $T_k \cdot T_{n+1-k}$ = constant = $a \cdot l$ (first × last).

2. Three terms in GP: If $a, b, c$ are in GP → $b^2 = ac$ (GM property).

3. Inserting $n$ GMs between $a$ and $b$: Common ratio $= (b/a)^{1/(n+1)}$.

4. GP with negative ratio: Terms alternate in sign.

5. Log of GP terms form AP: If $a_1, a_2, \ldots$ is GP, then $\log a_1, \log a_2, \ldots$ is AP.

❌ Mistake 1: Assuming the original sequence $\{a_n\}$ also has ratio $1/\sqrt{2}$. It does not — the ratio of $\{a_n\}$ is $\sqrt{2}$.

❌ Mistake 2: Computing $(\sqrt{2})^{10}$ incorrectly. Remember $(\sqrt{2})^{10} = 2^{10/2} = 2^5 = 32$, not $\sqrt{2}^{10} = 10\sqrt{2}$.

❌ Mistake 3: Using $S = a(1-r^n)/(1-r)$ with $r>1$. This gives negative denominator — use $a(r^n-1)/(r-1)$ when $r > 1$.

❌ Mistake 4: Not simplifying $\dfrac{62(\sqrt{2}-1)}{31} = 2(\sqrt{2}-1)$, and incorrectly rationalising the answer.

GP problems in JEE often involve:
• Modified sequences (like this one — divide/multiply terms by functions of $n$)
• Combined AP-GP problems (AGP — Arithmetico-Geometric Progression)
• Infinite GP with $|r|<1$
• Finding number of terms given first term, last term, ratio

For modified GP problems, always derive the ratio of the original sequence first before applying any sum formula. This is the most common source of error.

Time target: 2–3 minutes for this type in JEE Main.