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NCERT Solutions for Class 9th Maths Chapter 8 Exercise Set 8.1 — Recursive Rule for a Sequence

Book page 179–180 Updated on2026-09-08

Q1.
Find the first five terms of the sequence in which the nth term is given by (i) tn = 3n – 4, (ii) tn = 2 – 5n, and (iii) tn = n² – 2n + 3 for n ≥ 1.
Answer

Put n = 1, 2, 3, 4, 5 into each rule.

(i) tn = 3n – 4

t1 = 3 – 4 = –1   t2 = 6 – 4 = 2   t3 = 9 – 4 = 5   t4 = 12 – 4 = 8   t5 = 15 – 4 = 11
First five terms: –1, 2, 5, 8, 11

(ii) tn = 2 – 5n

t1 = 2 – 5 = –3   t2 = 2 – 10 = –8   t3 = 2 – 15 = –13   t4 = –18   t5 = –23
First five terms: –3, –8, –13, –18, –23

(iii) tn = n2 – 2n + 3

t1 = 1 – 2 + 3 = 2   t2 = 4 – 4 + 3 = 3   t3 = 9 – 6 + 3 = 6
t4 = 16 – 8 + 3 = 11   t5 = 25 – 10 + 3 = 18
First five terms: 2, 3, 6, 11, 18
Why it happens: the first two rules are linear in n, so their terms change by the same amount every step — by +3 in (i) and by –5 in (ii). Both are arithmetic progressions, one rising and one falling. The third rule is quadratic, so its gaps are 1, 3, 5, 7 — themselves growing — and it is not an AP. Reading the shape of the formula tells you the shape of the sequence before you compute a single term.
Check it yourself: for (iii), n2 – 2n + 3 = (n – 1)2 + 2, so the terms are 2 more than the square numbers 0, 1, 4, 9, 16 — giving 2, 3, 6, 11, 18 again.
Q2.
Find the 10th and 15th terms of the sequence tn = 5n – 3 for n ≥ 1.
Answer
t10 = 5 × 10 – 3 = 50 – 3 = 47
t15 = 5 × 15 – 3 = 75 – 3 = 72
Why it happens: tn = 5n – 3 is an AP with first term t1 = 2 and common difference 5. So the 15th term should sit five steps of 5 beyond the 10th: 47 + 5 × 5 = 72, which is exactly what the direct substitution gives. Two independent routes to the same value is a good check.
Q3.
Determine whether 97 and 172 are terms of the sequence tn = 5n – 3 for n ≥ 1.
Answer

Solve tn = 97 and tn = 172 and see whether n comes out a natural number.

5n – 3 = 97 → 5n = 100 → n = 20
5n – 3 = 172 → 5n = 175 → n = 35

Both values of n are natural numbers, so 97 is the 20th term and 172 is the 35th term — both are terms of the sequence.

Why it happens: every term of this sequence is 3 less than a multiple of 5, so it ends in 2 or 7. Both 97 and 172 end in 7 and 2, and adding 3 gives 100 and 175, each divisible by 5. Had we tested 98, we would have got 5n = 101 and n = 20.2 — not a position, so not a term.
Tip: "is k a term?" is always the same question in disguise: does the equation tn = k have a natural-number solution?
Q4.
Which term of the sequence tn = 5n – 3 for n ≥ 1 is 607?
Answer
5n – 3 = 607
5n = 610
n = 122

So 607 is the 122nd term. Check: 5 × 122 – 3 = 610 – 3 = 607. ✓

Why it happens: solving for n reverses the rule. The formula 5n – 3 takes a position and returns a value; the equation takes a value and returns its position. This two-way use is what makes an explicit rule so much more powerful than a list of terms.
Q5.
A sequence is given by the recursive rule t1 = – 5, tn+1 = tn + 3 for n ≥ 1. Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it?
Answer

Start at t1 = –5 and add 3 each time.

t2 = –5 + 3 = –2
t3 = –2 + 3 = 1
t4 = 1 + 3 = 4
t5 = 4 + 3 = 7
First five terms: –5, –2, 1, 4, 7

The terms rise by a constant 3, so this is an AP with a = –5 and d = 3. Convert to an explicit rule and then solve:

tn = –5 + (n – 1) × 3 = 3n – 8
3n – 8 = 52
3n = 60
n = 20

Yes — 52 is a term, the 20th term. Check: 3 × 20 – 8 = 52. ✓

Why it happens: the recursive rule alone would force you to compute nineteen terms before reaching 52. Turning it into the explicit rule tn = 3n – 8 first is the efficient move — and the two forms describe the same sequence, since t1 = 3 – 8 = –5 and each step in n adds 3.
Tip: the rule is written tn+1 = tn + 3 here, not tn = tn–1 + 3. Both say the same thing — the term after any term is 3 more — only the label has shifted.
Q6.
Let T1 = 1, T2 = 2, T3 = 4, and Tn = Tn–1 + Tn–2 + Tn–3 for n ≥ 4. Find T4, T5, T6, T7, and T8.
Answer

Each term is the sum of the three terms before it.

T4 = T3 + T2 + T1 = 4 + 2 + 1 = 7
T5 = T4 + T3 + T2 = 7 + 4 + 2 = 13
T6 = T5 + T4 + T3 = 13 + 7 + 4 = 24
T7 = T6 + T5 + T4 = 24 + 13 + 7 = 44
T8 = T7 + T6 + T5 = 44 + 24 + 13 = 81

The sequence is 1, 2, 4, 7, 13, 24, 44, 81, …

Why it happens: because the rule looks back three steps, it needs three starting terms, and it can only begin at n = 4. Notice how the pattern of the Virahānka rule extends: reaching back two terms gives 1, 2, 3, 5, 8, …; reaching back three gives 1, 2, 4, 7, 13, … The number of starting values a recursive rule needs is exactly the number of steps it reaches back.
Did you know? a sequence built by adding the previous three terms is called a tribonacci sequence, by analogy with the two-term rule.
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