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NCERT Solutions for Class 9th Maths Chapter 8 In-text Questions — Explicit Rule for a Sequence

Book page 177–178 Updated on2026-09-08

Q1.
Exercise: Using the explicit rule un = 2n – 1, find the 53rd term, the 108th term, and the 1170th term of the odd number sequence.
Answer

Substitute the position straight into the rule.

u53 = 2 × 53 – 1 = 106 – 1 = 105
u108 = 2 × 108 – 1 = 216 – 1 = 215
u1170 = 2 × 1170 – 1 = 2340 – 1 = 2339
Why it happens: the odd numbers are the even numbers shifted down by one. The nth even number is 2n, so the nth odd number is 2n – 1. Nothing here depends on knowing the earlier terms — this is exactly the advantage of an explicit rule, and it is why the 1170th term costs no more work than the 53rd.
Check it yourself: 105, 215 and 2339 are all odd, and each is one less than an even number that is twice its position.
Q2.
Here is the sequence of the first ten prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Do you see any pattern in this sequence? Can you think of a rule that can predict the next few prime numbers?
Answer

There are visible features, but no rule that predicts the next term. Compare the gaps:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29
gaps: 1, 2, 2, 4, 2, 4, 2, 4, 6

The gaps do not settle to a constant (so this is not an AP), and the ratios 3/2, 5/3, 7/5, … are not constant either (so it is not a GP). The next few primes — 31, 37, 41, 43, 47 — have to be found by testing for divisibility, not produced by a formula.

What can be said with certainty:

  • 2 is the only even prime; every later prime is odd.
  • Apart from 2 and 3, every prime is 1 more or 1 less than a multiple of 6.
  • The primes never run out — there are infinitely many of them.
Why it happens: a prime is defined by what it is not — it has no divisor other than 1 and itself. That is a condition about all the numbers below it, not a step forward from the previous term. So the definition gives you no way to move from one prime to the next, which is precisely why no explicit or recursive formula for the primes is known. This is the chapter's own example of a perfectly well-defined sequence with no clear regularity in its rule.
Did you know? the gaps between primes get arbitrarily large — somewhere beyond 100! there is a run of 99 consecutive numbers with no prime in it at all.
Q3.
Exercise: Consider the expression tn = 3n – 7. (i) Find its first, second, third, 12th, 18th and 50th terms. (ii) Which term of the sequence is 332? (iii) Is 557 a term of this sequence? Why or why not?
Answer

(i) Substitute each position into tn = 3n – 7.

t1 = 3 – 7 = –4    t2 = 6 – 7 = –1    t3 = 9 – 7 = 2
t12 = 36 – 7 = 29    t18 = 54 – 7 = 47    t50 = 150 – 7 = 143

(ii) Solve tn = 332.

3n – 7 = 332
3n = 339
n = 113

So 332 is the 113th term.

(iii) Solve tn = 557 the same way.

3n – 7 = 557
3n = 564
n = 564 ÷ 3 = 188

188 is a natural number, so yes — 557 is a term of the sequence, the 188th term. Check: 3 × 188 – 7 = 564 – 7 = 557.

Why it happens: every term of this sequence is 7 less than a multiple of 3, so a number k is a term exactly when k + 7 is divisible by 3 — that is, when k leaves remainder 2 on division by 3. Both 332 and 557 pass that test (332 = 3 × 110 + 2 and 557 = 3 × 185 + 2), so both belong. A number such as 558, which leaves remainder 0, could never appear however far you continued.
Tip: a "why or why not" question is really asking for the test, not just the verdict. Here the test is: does the equation 3n – 7 = k give a natural number n?
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