Q.Which term of the following sequences:

Concept: Geometric Progression (GP)

For a GP with first term $a$ and common ratio $r$, the $n$th term is $a r^{n-1}$.

(a) $2, 2\sqrt{2}, 4, \ldots$

$a = 2$, $r = \sqrt{2}$.

Set $a r^{n-1} = 128$:

$2 (\sqrt{2})^{n-1} = 128 \implies (\sqrt{2})^{n-1} = 64 = 2^6 = (\sqrt{2})^{12}$.

So $n-1 = 12 \implies n = 13$.

(b) $\sqrt{3}, 3, 3\sqrt{3}, \ldots$

$a = \sqrt{3}$, $r = \sqrt{3}$.

$\sqrt{3} (\sqrt{3})^{n-1} = 729 = 3^6 = (\sqrt{3})^{12}$.

Thus $n = 12$.

(c) $\dfrac{1}{3}, \dfrac{1}{9}, \dfrac{1}{27}, \ldots$

$a = \frac{1}{3}$, $r = \frac{1}{3}$. …

A geometric progression (GP) is a sequence where each term after the first is obtained by multiplying the previous term by a fixed number called the common ratio ($r$). The key idea: if you know the first term $a_1$ and the ratio $r$, any term can be written as $a_n = a_1 r^{n-1}$. So to find which term a given number is, we set up that equation and solve for $n$ — which usually involves expressing both sides as powers of the same base.

Let’s work through each part.

(a) $2, 2\sqrt{2}, 4, \ldots$ is $128$?

1. Identify the first term and common ratio.

First term $a_1 = 2$.

Second term is $2\sqrt{2}$. The ratio $r = \frac{2\sqrt{2}}{2} = \sqrt{2}$.

Check: third term $2 \times (\sqrt{2})^2 = 2 \times 2 = 4$, matches.

2. Write the general term.

$a_n = 2 \cdot (\sqrt{2})^{n-1}$.

3. Set equal to $128$ and solve for $n$.

$2 \cdot (\sqrt{2})^{n-1} = 128$

Divide both sides by $2$: $(\sqrt{2})^{n-1} = 64$.

Now express $64$ as a power of $\sqrt{2}$. Since $\sqrt{2} = 2^{1/2}$, we have $(\sqrt{2})^{n-1} = 2^{(n-1)/2}$. And $64 = 2^6$.

So $2^{(n-1)/2} = 2^6$. Equating exponents: $\frac{n-1}{2} = 6 \implies n-1 = 12 \implies n = 13$.

4. Verify: $a_{13} = 2 \cdot (\sqrt{2})^{12} = 2 \cdot 64 = 128$. Correct.

(b) $\sqrt{3}, 3, 3\sqrt{3}, \ldots$ is $729$?

1. First term and ratio.

$a_1 = \sqrt{3}$.

$r = \frac{3}{\sqrt{3}} = \sqrt{3}$ (since $3 = \sqrt{3} \cdot \sqrt{3}$).

Check: third term $\sqrt{3} \cdot (\sqrt{3})^2 = \sqrt{3} \cdot 3 = 3\sqrt{3}$, matches.

2. General term: $a_n = \sqrt{3} \cdot (\sqrt{3})^{n-1} = (\sqrt{3})^n$.

3. Set equal to $729$: $(\sqrt{3})^n = 729$.

Write $\sqrt{3} = 3^{1/2}$, so $(\sqrt{3})^n = 3^{n/2}$. And $729 = 3^6$ (since $3^6 = 729$).

Thus $3^{n/2} = 3^6 \implies \frac{n}{2} = 6 \implies n = 12$.

4. Check: $a_{12} = (\sqrt{3})^{12} = 3^{6} = 729$. Good.

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