Exercise Set 8.1 - Infinity Hub

Class 9 Mathematics Ganita Manjari Part 1 NCERT Solutions
Chapter-8: Predicting What Comes Next: Exploring Sequences and Progressions

Exercise Set 8.1

1. Find the first five terms of the sequence in which the $n^\text{th}$ term is given by (i) $t_n=3n-4$ (ii) $t_n=2-5n$ (iii) $t_n=n^2 -2n+3$ for $n \ge 1$ .

Solution:
(i) $t_n=3n-4$
Put $n=1,2,3,4,5$
$t_1=3(1)-4=3-4=-1$
$t_2=3(2)-4=6-4=2$
$t_3=3(3)-4=9-4=5$
$t_4=3(4)-4=12-4=8$
$t_5=3(5)-4=15-4=11$
So, first five terms are -1, 2, 5, 8, 11

(ii) $t_n=2-5n$
Put $n=1,2,3,4,5$
$t_1=2-5(1)=2-5=-3$
$t_2=2-5(2)=2-10=-8$
$t_3=2-5(3)=2-15=-13$
$t_4=2-5(4)=2-20=-18$
$t_5=2-5(5)=2-25=-23$
So, first five terms are -3, -8, -13, -18, -23

(iii) $t_n=n^2 -2n+3$
Put $n=1,2,3,4,5$
$t_1=1^2 -2(1)+3=1-2+3=2$
$t_2=2^2 -2(2)+3=4-4+3=3$
$t_3=3^2 -2(3)+3=9-6+3=6$
$t_4=4^2 -2(4)+3=16-8+3=11$
$t_5=5^2 -2(5)+3=25-10+3=18$
So, first five terms are 2,3, 6, 11, 18

2. Find the $10^\text{th}$ and $15^\text{th}$ terms of the sequence $t_n=5n-3$ for $n \ge 1$ .

Solution: $t_n=5n-3$
$10^\text{th}$ term (n=10)
$t_{10}=5(10)-3$
$=50-3$
$=47$
$15^\text{th}$ term (n=15)
$t_{15}=5(15)-3$
$=75-3$
$=72$

3. Determine whether 97 and 172 are terms of the sequence $t_n=5n-3$ for $n \ge 1$ .

Solution: $t_n=5n-3$
If 97 is a term of given sequence, then
$t_n=97$
$5n-3=97$
$5n=97+3$
$5n=100$
$n=\frac{100}{5}=20$
Since $n = 20$ is a positive integer, 97 is a term of given sequence.

If 172 is a term of given sequence, then
$t_n=172$
$5n-3=172$
$5n=172+3$
$5n=175$
$n=\frac{175}{5}=35$
Since $n =35$ is a positive integer, 172 is a term of given sequence.

4. Which term of the sequence $t_n=5n-3$ for $n \ge 1$ is 607.

Solution: $t_n=5n-3$
Let $t_n=607$
$5n-3=607$
$5n=607+3$
$5n=610$
$n=\frac{610}{5}=122$
607 is the 122nd term of given sequence

5.A sequence is given by the recursive rule $t_1=-5, t_{n+1}=t_n+3$ for $n \ge 1$ . Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it?

Solution:
Given: $t_1=-5, t_{n+1}=t_n+3$
Put $n=1$
$t_{1+1}=t_1+3$
$t_2=-5+3=-2$
Put $n=2$
$t_{2+1}=t_2+3$
$t_3=-2+3=1$
Put $n=3$
$t_{3+1}=t_3+3$
$t_4=1+3=4$
Put $n=4$
$t_{4+1}=t_4+3$
$t_5=4+3=7$
Thus, first five terms are -5, -2, 1, 4, 7
This is an arithmetic sequence with first term -5 and common difference 3
So, its $n^{th}$ term is
$t_n=-5+(n-1)3$
$t_n=-5+3n-3$
$t_n=3n-8$
Let $t_n=52$
$3n-8=52$
$3n=52+8$
$3n=60$
$n=\frac{60}{3}$
$n=20$
Therefore, 52 is the 20th term of the given sequence.

6. Let $T_1=1, T_2=2, T_3=4$ , and $T_n=T_{n-1}+T_{n-2}+T_{n-3}$ for $n \ge 4$ . Find $T_5, T_6, T_7,$ and $T_8$ .

Solution:
Given: $T_1=1, T_2=2, T_3=4$
$T_n=T_{n-1}+T_{n-2}+T_{n-3}$
Put $n=4$
$T_4=T_{4-1}+T_{4-2}+T_{4-3}$
$T_4=T_3+T_2+T_1$
$T_4=4+2+1$
$T_4=7$
Put $n=5$
$T_5=T_{5-1}+T_{5-2}+T_{5-3}$
$T_5=T_4+T_3+T_2$
$T_5=7+4+2$
$T_5=13$
Put $n=6$
$T_6=T_{6-1}+T_{6-2}+T_{6-3}$
$T_6=T_5+T_4+T_3$
$T_6=13+7+4$
$T_6=24$
Put $n=7$
$T_7=T_{7-1}+T_{7-2}+T_{7-3}$
$T_7=T_6+T_5+T_4$
$T_7=24+13+7$
$T_7=44$
Put $n=8$
$T_8=T_{8-1}+T_{8-2}+T_{8-3}$
$T_8=T_7+T_6+T_5$
$T_8=44+24+13$
$T_8=81$