The lengths of three unequal edges of a rectangular solid block are in G.P. The volume of the block is $216$ cm$^3$ and the total surface area is $252$ cm$^2$. The length of the longest edge is (A) $12$ cm (B) $6$ cm (C) $18$ cm (D) $3$ cm

The lengths of three unequal edges of a rectangular solid block are in G.P. The volume of the block is 216 cm³

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Concept understanding — Geometric Progression

Geometric Progression: The Idea of Repeated Multiplication

Imagine you're folding a piece of paper in half. Start with thickness 1 unit. After one fold, thickness becomes 2. After two folds, thickness becomes 4. After three folds, thickness becomes 8. The sequence of thicknesses is:

1, 2, 4, 8, 16, ...

Notice the pattern: each term is obtained by multiplying the previous term by the same number (here, 2). That's the core intuition behind a geometric progression — you keep multiplying by a fixed number, step after step.

This is different from an arithmetic progression, where you keep adding a fixed number. Here, the growth is multiplicative, not additive. That's why geometric progressions grow (or shrink) much faster.

Precise Definition

A Geometric Progression (GP) is a sequence of numbers where the ratio of any term to its preceding term is constant. This constant is called the common ratio, denoted by $r$ .

If the first term is $a$ , then the sequence looks like:

a, \ ar, \ ar^2, \ ar^3, \ ar^4, \ \dots

Note

The common ratio $r$ can be any real number — positive, negative, or even a fraction. If $r$ is negative, the terms alternate in sign. If $0 < r < 1$ , the terms get smaller and smaller.

The $n$ -th Term

To find any term directly without listing all previous ones, use the formula:

T_n = a \cdot r^{\,n-1}

where $T_n$ is the $n$ -th term, $a$ is the first term, $r$ is the common ratio, and $n$ is the term number (starting from 1).

Example: For the paper-folding sequence, $a = 1$ , $r = 2$ . The 5th term is $1 \cdot 2^{5-1} = 2^4 = 16$ , which matches our list.

Sum of $n$ Terms

There are two cases, depending on whether $r = 1$ or not.

n

When $r = 1$ , every term is just $a$ , so the sum is simply $n \times a$ .

Why the formula works (intuition):

Let $S = a + ar + ar^2 + \dots + ar^{n-1}$ . Multiply both sides by $r$ : $rS = ar + ar^2 + \dots + ar^n$ . Subtract the first from the second: $rS - S = ar^n - a$ , so $S(r-1) = a(r^n - 1)$ , giving the formula above.

Sum of an Infinite GP

If the common ratio $r$ lies strictly between $-1$ and $1$ (i.e., $|r| < 1$ ), the terms get smaller and smaller, and the sum of all terms approaches a finite value:

S_\infty = \frac{a}{1 - r}, \quad \text{for } |r| < 1

Watch out

If $|r| \geq 1$ , the infinite sum does not exist (it diverges to infinity or oscillates without settling). Never apply the infinite sum formula when $|r| \geq 1$ .

Example: $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$ has $a=1$ , $r=\frac12$ , so $S_\infty = \frac{1}{1 - 1/2} = 2$ . This matches the intuition that repeatedly halving a unit length eventually fills exactly 2 units.

Quick Reference Table

Property	Formula	Condition
Common ratio	$r = \dfrac{T_{n+1}}{T_n}$	Always
$n$ -th term	$T_n = a r^{n-1}$	Always

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The three edge lengths are in geometric progression, so we let them be $a/r$ , $a$ , $ar$ . Their product (volume) gives $a^3 = 216$ , so $a = 6$ . The surface area equation then yields $r = 2$ or $r = 1/2$ . The longest edge is $ar = 12$ cm, which corresponds to option (A).

The problem gives us a rectangular solid (a box) whose three edge lengths — let’s call them $x$ , $y$ , $z$ — are in geometric progression. That means there is a common ratio $r$ such that each term is the previous one multiplied by $r$ . The volume is $216$ cm $^3$ and the total surface area is $252$ cm $^2$ . We need the longest edge.

Why does letting the edges be $a/r$ , $a$ , $ar$ work so neatly? Because in a GP, the middle term acts as the geometric mean of the other two. If we set the middle edge to $a$ , then the smaller is $a/r$ and the larger is $ar$ . This keeps the product simple: $(a/r) \cdot a \cdot (ar) = a^3$ . That’s the key — the volume becomes just $a^3$ , which is immediately solvable.

Let’s walk through it.

Set up the edges in GP.

Let the three edge lengths be $\frac{a}{r}$ , $a$ , and $ar$ , where $a > 0$ and $r > 0$ . (If $r < 1$ , the order flips, but the longest edge will still be the one with the largest magnitude — we’ll handle that later.)
Use the volume.

Volume $V = \left(\frac{a}{r}\right) \cdot a \cdot (ar) = a^3$ .

Given $V = 216$ , we have

a^3 = 216 \quad\Rightarrow\quad a = \sqrt[3]{216} = 6.

So the middle edge is $6$ cm. The three edges are $\frac{6}{r}$ , $6$ , and $6r$ .

Use the total surface area. Surface area of a rectangular solid: $2(xy + yz + zx)$ . Here:

x = \frac{6}{r},\quad y = 6,\quad z = 6r.

Compute each product:

xy = \frac{6}{r} \cdot 6 = \frac{36}{r},

yz = 6 \cdot 6r = 36r,

zx = 6r \cdot \frac{6}{r} = 36.

So total surface area:

2\left(\frac{36}{r} + 36r + 36\right) = 252.

Simplify the equation. Divide both sides by $2$ :

\frac{36}{r} + 36r + 36 = 126.

Subtract $36$ :

\frac{36}{r} + 36r = 90.

Divide through by $18$ (or by $36$ — either works):

\frac{2}{r} + 2r = 5.

Multiply through by $r$ :

2 + 2r^2 = 5r \quad\Rightarrow\quad 2r^2 - 5r + 2 = 0.

Solve the quadratic. …

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Q.The lengths of three unequal edges of a rectangular solid block are in G.P. The volume of the block is 216216216 cm3^33 and the total surface area is 252252252 cm2^22. The length of the longest edge is (A) 121212 cm (B) 666 cm (C) 181818 cm (D) 333 cm

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Q.The lengths of three unequal edges of a rectangular solid block are in G.P. The volume of the block is $216$ cm $^3$ and the total surface area is $252$ cm $^2$ . The length of the longest edge is
(A) $12$ cm
(B) $6$ cm
(C) $18$ cm
(D) $3$ cm

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