Which term of the following sequences: (i) 5, 10, 20, 40, … is 5120 (ii) 2, 2√(2), 4, … is 128 (iii) 2, 1, (1)

Exercise 5.2 · Q2

Q. Which term of the following sequences:

(i)

5, 10, 20, 40, \ldots

is 5120

(ii)

2, 2\sqrt{2}, 4, \ldots

is 128

(iii)

2, 1, \dfrac{1}{2}, \dfrac{1}{4}, \ldots

\dfrac{1}{128}

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Concept understanding — Sequences

Sequences: From Intuition to Definition

Think of a sequence as a numbered list that goes on forever. You already know lists — a grocery list, a to-do list. A sequence is just that, but with two special rules: every item gets a natural number label (1st, 2nd, 3rd, …), and the list never ends.

For example, the odd numbers: 1, 3, 5, 7, 9, … That’s a sequence. The 1st term is 1, the 2nd is 3, the 3rd is 5, and so on. The "…" means it keeps going — there is no last term.

The key intuition: a sequence is an ordered, infinite parade of numbers, where each number has a fixed position.

The Precise Definition

A sequence is a function whose domain is the set of natural numbers $\mathbb{N} = \{1, 2, 3, \dots\}$ (or sometimes $\{0, 1, 2, \dots\}$ ) and whose range is a set of real numbers.

We write the sequence as $\{a_n\}_{n=1}^\infty$ or simply $(a_n)$ , where $a_n$ is the $n$ th term of the sequence.

Important

A sequence is not a set. Order matters, and repetition is allowed. The sequence $1, 2, 1, 2, 1, 2, \dots$ is perfectly valid, even though the set $\{1, 2\}$ has only two elements.

How We Describe a Sequence

There are three common ways to specify a sequence:

1. Listing the first few terms — enough to show the pattern.

Example: $2, 4, 6, 8, 10, \dots$ (the even numbers).

2. A formula for the $n$ th term — the most powerful method.

Example: $a_n = 2n$ . Then $a_1 = 2$ , $a_2 = 4$ , $a_3 = 6$ , and so on.

3. A recurrence relation — each term is defined using previous terms, plus a starting value.

Example: $a_1 = 1$ , and $a_n = a_{n-1} + 2$ for $n \ge 2$ . This also gives $1, 3, 5, 7, \dots$ .

Tip

When you see a sequence like $1, 4, 9, 16, \dots$ , ask: "What is the pattern in terms of the position $n$ ?" Here $a_n = n^2$ .

Why This Matters

Sequences are the foundation of limits and series — the core of calculus. Every time you sum an infinite list (like $1 + \frac12 + \frac14 + \frac18 + \cdots$ ), you are working with a sequence of partial sums. Understanding sequences first makes everything that follows — convergence, divergence, infinite sums — crystal clear.

Note

In Indian exams (CBSE, JEE, etc.), you will encounter arithmetic progressions, geometric progressions, and harmonic progressions as special types of sequences. But every sequence is just a function from $\mathbb{N}$ to $\mathbb{R}$ — that is the single idea that unifies them all.

Quick Check

Which of these is a sequence?

$\{2, 4, 6, 8\}$
$2, 4, 6, 8, \dots$
$a_n = \frac{1}{n}$ for $n = 1, 2, 3, \dots$ Answer: (ii) and (iii) are sequences. (i) is just a finite set — no "…" and no infinite continuation.

Solving $a_n=ar^{n-1}$ for each sequence: $5120$ is the 11th term, $128$ is the 13th term, and $\dfrac1{128}$ is the 9th term.

n

(i) $5,10,20,40,\ldots$ ; which term is $5120$ ?

$a=5$ , $r=\dfrac{10}{5}=2$ .
$ar^{n-1}=5120\Rightarrow 5(2)^{n-1}=5120\Rightarrow 2^{n-1}=1024=2^{10}$ .
$n-1=10\Rightarrow n=11$ .

(ii) $2,2\sqrt2,4,\ldots$ ; which term is $128$ ?

4. $a=2$ , $r=\dfrac{2\sqrt2}{2}=\sqrt2$ (check: $\dfrac{4}{2\sqrt2}=\sqrt2$ ✓).

5. $2(\sqrt2)^{n-1}=128\Rightarrow (\sqrt2)^{n-1}=64=2^6\Rightarrow 2^{\frac{n-1}{2}}=2^6\Rightarrow \dfrac{n-1}{2}=6\Rightarrow n-1=12\Rightarrow n=13$ .

(iii) $2,1,\dfrac12,\dfrac14,\ldots$ ; which term is $\dfrac1{128}$ ?

6. $a=2$ , $r=\dfrac12$ .

7. $2\left(\dfrac12\right)^{n-1}=\dfrac1{128}\Rightarrow \left(\dfrac12\right)^{n-1}=\dfrac1{256}=\left(\dfrac12\right)^8\Rightarrow n-1=8\Rightarrow n=9$ .

8. Self-check (iii): $a_9=2\left(\dfrac12\right)^8=\dfrac{2}{256}=\dfrac1{128}$ . ✓

✓Final answer

(i) 11th term (ii) 13th term (iii) 9th term

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