Write the first five terms of the sequence whose nth term is aₙ = (-1)ⁿ⁻¹5ⁿ⁺¹.

Exercise 8.1 · Q5

n

Karnataka PUCTextbookSubjective· 2mImportance★★★★★est

4% · 5/114 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Sequence Terms Evaluation

Sequence Terms Evaluation

A parking garage numbers its levels 1, 2, 3, ... and posts a rule on the wall: "Level $n$ sits $n^2$ metres below street level." You don't need to walk down to level 10 to find its depth -- you simply compute $10^2 = 100$ metres. That one substitution is the entire idea behind sequence terms evaluation: once a sequence has a rule connecting position to value, any term can be found directly, without listing every term before it.

What "Evaluating a Term" Means

A sequence assigns a number to each position $1, 2, 3, \dots$ . The term at position $n$ is written $a_n$ (read "a sub n"). When the sequence is given by an explicit formula $a_n = f(n)$ , evaluating the $k$ -th term simply means substituting $n = k$ into $f$ and simplifying.

Note

This is different from a recursive definition, where $a_n$ is defined using $a_{n-1}$ -- for example, $a_1 = 1,\ a_n = a_{n-1} + 2$ . A recursive rule forces you to compute every earlier term first; an explicit formula lets you jump straight to any term you want.

The Precise Statement

If $\{a_n\}$ is a sequence with $a_n = f(n)$ , then for any positive integer $k$ , the $k$ -th term is

a_k = f(k)

The entire procedure is: replace $n$ with the required index, then simplify the resulting number. No equation-solving, no guesswork -- just substitution.

Common Forms You'll Meet in Exams

Type	Formula	5th term
Linear	$a_n = 3n - 2$	$3(5)-2 = 13$
Quadratic	$a_n = n^2 + 1$	$5^2+1 = 26$
Exponential	$a_n = 2^n$	$2^5 = 32$
Rational	$a_n = \dfrac{n}{n+1}$	$\dfrac{5}{6}$
Alternating	$a_n = (-1)^n$	$(-1)^5 = -1$

Worked Example

Find the 7th term of the sequence $a_n = \dfrac{3n-1}{n^2+2}$ .

The index we need is $k = 7$ .
Substitute $n = 7$ :

a_7 = \frac{3(7)-1}{7^2+2} = \frac{21-1}{49+2} = \frac{20}{51}

Check whether it simplifies -- $\gcd(20, 51) = 1$ , so $\dfrac{20}{51}$ is already in lowest terms and is the final answer. …

Start your 14-day free trial to unlock the full solution →

The sequence alternates signs because of $(-1)^{n-1}$ , and each term grows by a factor of 5. The first five terms are $25, -125, 625, -3125, 15625$ .

The key here is to see the pattern in two parts: the sign and the magnitude. The factor $(-1)^{n-1}$ controls whether the term is positive or negative, while $5^{n+1}$ gives the size. When $n$ is odd, $n-1$ is even, so $(-1)^{\text{even}} = 1$ — the term is positive. When $n$ is even, $n-1$ is odd, so the term is negative. The exponent $n+1$ starts at $2$ when $n=1$ , then increases by 1 each step, so the magnitudes are $5^2, 5^3, 5^4, \dots$ .

Let’s write them out step by step.

$n = 1$ :

$a_1 = (-1)^{1-1} \cdot 5^{1+1} = (-1)^0 \cdot 5^2 = 1 \cdot 25 = 25$ .
$n = 2$ :

$a_2 = (-1)^{2-1} \cdot 5^{2+1} = (-1)^1 \cdot 5^3 = -1 \cdot 125 = -125$ .
$n = 3$ :

$a_3 = (-1)^{3-1} \cdot 5^{3+1} = (-1)^2 \cdot 5^4 = 1 \cdot 625 = 625$ .
$n = 4$ :

$a_4 = (-1)^{4-1} \cdot 5^{4+1} = (-1)^3 \cdot 5^5 = -1 \cdot 3125 = -3125$ .
$n = 5$ : …

Start your 14-day free trial to unlock the full solution →

←Previous question · 8.3Q4 — Write the first five terms of the sequence whose nth term is aₙ = (2n-…→Next question · 8.3Q6 — Write the first five terms of the sequence whose nth term is aₙ = n(n²…

Q.Write the first five terms of the sequence whose nnnth term is an=(−1)n−1 5n+1a_n = (-1)^{n-1}\,5^{n+1}an​=(−1)n−15n+1.

🎓Unlock everything free for 14 days

Q.Write the first five terms of the sequence whose $n$ th term is $a_n = (-1)^{n-1}\,5^{n+1}$ .

Unlock everything free for 14 days