Class 9 Mathematics Ganita Manjari Part 1 NCERT Solutions
Chapter-8: Predicting What Comes Next: Exploring Sequences and Progressions
Exercise Set 8.2
1. Find the 10^{th} and 26^{th} terms of the AP: 3, 8, 13, 18, ….
Solution:
Given AP : 3, 8, 13, 18…
First term, a=3
Common difference, d= 8-3=5
The n^{th} term of an AP is:
a_n=a+(n-1)d
10^{th} term =a_{10}
a_{10}=3+(10-1)5
a_{10}=3+9×5
a_{10}=3+45
a_{10}=48
26^{th} term =a_{26}
a_{26}=3+(26-1)5
a_{26}=3+25×5
a_{26}=3+125
a_{26}=128
2. Which term of the AP : 21, 18, 15, … is – 81? Also, is 0 a term of this AP? Give reasons for your answer.
Solution:
Given AP : 21, 18, 15…
First term, a=21
Common difference, d=18-21=-3
The n^{th} term of an AP is:
a_n=a+(n-1)d
Let a_n=-81
21+(n-1)(-3)=-81
21-3n+3=-81
24-3n=-81
3n=24+81
3n=105
n=\frac{105}{3}
n=35
Therefore, -81 is the 35^{th} term of given AP.
Now if 0 is a term of given AP, then
a_n=0
21+(n-1)(-3)=0
21-3n+3=0
24-3n=0
3n=24
n=\frac{24}{3}
n=8
Therefore, 0 is the 8^{th} term of given AP.
3. Find the n^{th} term of the AP: 11, 8, 5, 2 … Write the recursive rule for this AP.
Solution:
Given AP : 11, 8, 5, 2…
a=11
d=8-11=-3
n^{th} term is:
a_n=a+(n-1)d
a_n=11+(n-1)(-3)
a_n=11-3n+3
a_n=14-3n
Recursive Rule:
a_1=11
a_n=a_{n-1}-3, n \ge 2
4. An AP consists of 50 terms in which the 3^{rd} term is 12 and the last term is 106. Find the 29th term.
Solution:
Given: n=50, a_3=12, a_{50}=106
Let a is the first term and d is the common difference.
a_3=12
a+2d=12…………….(1)
a_{50}=106
a+49d=106…………….(2)
Subtract (1) from (2)
a+49d-(a+2d)=106-12
a+49d-a-2d=94
47d=94
d=\frac{94}{47}
d=2
Putting d=2 in (1)
a+2×2=12
a+4=12
a=12-4
a=8
Now, a_{29}=a+(n-1)d
a_{29}=8+(29-1)2
a_{29}=8+28×2
a_{29}=8+56
a_{29}=64
5. How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?
Solution:
Two digit numbers divisible by 3 are:
12, 15, 18,…., 99
So, First term,a=12
Common difference,d=3
Last term,l=99
Now, l=a+(n-1)d
99=12+(n-1)3
99-12=(n-1)3
87=(n-1)3
\frac{87}{3}=n-1
29=n-1
29+1=n
n=30
So, there are 30 two digit numbers divisible by 3.
Sum, S_n=\frac{n}{2}(a+l)
S_{30}=\frac{30}{2}(12+99)
S_{30}=15(111)
S_{30}=1665
Thus, sum of all two digit numbers divisible by 3 is 1665.
6. Harish started work at an annual salary of ₹5,00,000 and received an increment of ₹20,000 each year. After how many years did his income reach ₹7,00,000?
Solution:
Initial Salary= ₹5,00,000
Annual increment= ₹20,000
Salary after first increment= ₹5,00,000 + ₹20,000= ₹5,20,000
Salary after second increment= ₹5,20,000 + ₹20,000= ₹5,40,000
So, salary forms an AP:
500000, 520000, 540000,….
Here,
a = 5,00,000
d = 20,000
Let income become ₹7,00,000 in n^{th} year
a_n=7,00,000
a+(n-1)d=7,00,000
5,00,000+(n-1)20,000=7,00,000
(n-1)20,000=7,00,000-5,00,000
(n-1)20,000=2,00,000
n-1=\frac{2,00,000}{20,000}
n-1=10
n=10+1
n=11
So, ₹7,00,000 is his 11^{th} year’s salary
Therefore, his income reached ₹7,00,000 after 10 years.
7. A child arranges marbles in rows so that the first row has 1 marble, the second has 2 marbles, the third has 3, and so on up to 25 rows. How many marbles does the child use in all?
Solution:
Marbles in first row = 1
Marbles in second row = 2
Marbles in third row = 3
This forms an AP: 1, 2, 3,… up to 25
a=1, d=1, n=25
Sum, s_n=\frac{n}{2}[2a+(n-1)d]
s_{25}=\frac{25}{2}[2×1+(25-1)1]
s_{25}=\frac{25}{2}[2+(24)1]
s_{25}=\frac{25}{2}×26
s_{25}=25×13
s_{25}=325
Therefore, the child uses 325 marbles.