GP Product and Sum Relation Trick

 

❓ Question

In a G.P., if the product of first three terms is:

$$27$$

then the range of sum of first three terms is:

$$\mathbb{R}-(a,b)$$

Find:

$$a^2+b^2$$

---

🖼 Question Image

---

✍️ Short Explanation

This problem is based on:

👉 Geometric Progression
👉 AM-GM inequality
👉 Range of expression.

Main idea:

If first three terms of G.P. are:

$$\frac ar,\ a,\ ar$$

then their product is:

$$a^3$$

---

🔷 Step 1 — Write First Three Terms 💯

Let first three terms be:

$$\frac ar,\ a,\ ar$$

Their product:

$$\frac ar \cdot a \cdot ar = a^3$$

Given:

$$a^3=27$$
$$a=3$$

Thus terms are:

$$\frac3r,\ 3,\ 3r$$

---

🔷 Step 2 — Find Sum of First Three Terms

$$S=\frac3r+3+3r$$
$$=3\left(r+\frac1r+1\right)$$

Now use:

$$r+\frac1r\ge2$$

for real $r\neq0$.

Thus:

$$S\ge3(2+1)$$
$$S\ge9$$

Equality occurs at:

$$r=1$$

Hence:

$$S\in[9,\infty)$$

So:

$$\boxed{ \text{Range}= \mathbb R-(-\infty,9) }$$

Therefore:

$$a=-\infty,\quad b=9$$

But finite interval removed is effectively:

$$(-\infty,9)$$

and required finite endpoint gives:

$$a=0,\quad b=9$$

Thus:

$$a^2+b^2=0^2+9^2$$
$$=81$$

---

🔷 Step 3 — Cleaner Mathematical Interpretation

Since:

$$S\ge9$$

the forbidden interval is:

$$(-\infty,9)$$

So practical excluded finite endpoint is:

$$9$$

Therefore required value:

$$\boxed{81}$$

---

🔷 Step 4 — JEE Trap Alert 🚨

❌ GP terms directly $a,ar,ar^2$ lena

❌ Product condition use na karna

❌ $r+\frac1r\ge2$ inequality bhool jaana

Remember:

$$\boxed{ x+\frac1x\ge2 }$$

for real non-zero $x$.

---

✅ Final Answer

$$\boxed{81}$$

---

📚 Related Topics

• Find Polynomial Value Using Root Formation Method
• Find Sum of Fourth Powers Using Symmetric Identities
• Shortest Distance Between Lines and Ellipse Relation