Mathematics solution NCERT
Class 9 - Chapter 8: Predicting What Comes Next: Exploring Sequences and Progressions
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.
(i) tn = 3n − 4
Substituting n = 1, 2, 3, 4, 5:
| n | tn = 3n − 4 |
|---|---|
| 1 | 3(1) − 4 = −1 |
| 2 | 3(2) − 4 = 2 |
| 3 | 3(3) − 4 = 5 |
| 4 | 3(4) − 4 = 8 |
| 5 | 3(5) − 4 = 11 |
First Five Terms: −1, 2, 5, 8, 11
(ii) tn = 2 − 5n
| n | tn = 2 − 5n |
|---|---|
| 1 | 2 − 5 = −3 |
| 2 | 2 − 10 = −8 |
| 3 | 2 − 15 = −13 |
| 4 | 2 − 20 = −18 |
| 5 | 2 − 25 = −23 |
First Five Terms: −3, −8, −13, −18, −23
(iii) tn = n2 − 2n + 3
| n | tn = n2 − 2n + 3 |
|---|---|
| 1 | 12 − 2(1) + 3 = 2 |
| 2 | 22 − 2(2) + 3 = 3 |
| 3 | 32 − 2(3) + 3 = 6 |
| 4 | 42 − 2(4) + 3 = 11 |
| 5 | 52 − 2(5) + 3 = 18 |
First Five Terms: 2, 3, 6, 11, 18
Q2: Find the 10th and 15th terms of the sequence tn = 5n – 3 for n ≥ 1.
Given: tn = 5n − 3
10th Term:
t10 = 5(10) − 3
= 50 − 3
= 47
Answer: t10 = 47
15th Term:
t15 = 5(15) − 3
= 75 − 3
= 72
Answer: t15 = 72
Q3: Determine whether 97 and 172 are terms of the sequence tn = 5n – 3 for n ≥ 1.
Given: tn = 5n − 3
Checking 97
5n − 3 = 97
5n = 100
n = 20
Since n is a whole number, 97 is the 20th term.
Checking 172
5n − 3 = 172
5n = 175
n = 35
Since n is a whole number, 172 is the 35th term.
Q4: Which term of the sequence tn = 5n – 3 for n ≥ 1 is 607?
Given: tn = 5n − 3
5n − 3 = 607
5n = 610
n = 122
Answer: 607 is the 122nd term of the sequence.
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?
Given: t1 = −5
tn+1 = tn + 3
First Five Terms
| Term | Value |
|---|---|
| t1 | −5 |
| t2 | −2 |
| t3 | 1 |
| t4 | 4 |
| t5 | 7 |
First Five Terms: −5, −2, 1, 4, 7
Is 52 a Term?
This is an arithmetic sequence with:
a = −5, d = 3
tn = a + (n − 1)d
52 = −5 + (n − 1)3
57 = 3(n − 1)
19 = n − 1
n = 20
Answer: Yes, 52 is the 20th term.
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 .
T1 = 1, T2 = 2, T3 = 4
Tn = Tn−1 + Tn−2 + Tn−3 for n ≥ 4
| Term | Calculation | Value |
|---|---|---|
| T4 | 4 + 2 + 1 | 7 |
| T5 | 7 + 4 + 2 | 13 |
| T6 | 13 + 7 + 4 | 24 |
| T7 | 24 + 13 + 7 | 44 |
| T8 | 44 + 24 + 13 | 81 |
Answers:
T4 = 7
T5 = 13
T6 = 24
T7 = 44
T8 = 81