Class 9 Mathematics Ganita Manjari Part 1 NCERT Solutions
Chapter-8: Predicting What Comes Next: Exploring Sequences and Progressions
End-of-Chapter Excersice
1. Find the 31^{st} term of an AP whose 11^{th} term is 38 and 16^{th} term is 73.
Solution:
Given a_{11}=38
⇒ a+10d=38 …….. (1)
Also, a_{16}=73
⇒ a+15d=73 …….. (2)
Subtract (1) from (2)
(a+15d)−(a+10d)=73−38
a+15d−a-10d=35
5d=35
d=\frac{35}{5}
d=7
Put d=7 in (1)
a+10(7)=38
a+70=38
a=38-70
a=-32
Now, 31^{st} term:
a_{31}=a+30d
a_{31}=-32+30(7)
a_{31}=-32+210
a_{31}=178
2. Determine the AP whose third term is 16 and whose 7^{th} term exceeds the 5^{th} term by 12.
Solution:
Given a_3=16
a+2d=16 ……. (1)
Also, 7^{th} term exceeds the 5^{th} term by 12
⇒ a_7=a_5+12
a+6d=a+4d+12
a+6d-a-4d=12
2d=12
d=\frac{12}{2}
d=6
Put d=6 in (1)
a+2(6)=16
a+12=16
a=16-12
a=4
Hence, First term, a=4 and Common difference, d=6
So, the AP is 4,10,16,22,28,…
*3. How many three-digit numbers are divisible by 7?
Solution:
Smallest three digit number divisible by 7 = 105
Largest three digit number divisible by 7 = 994
Three digit numbers divisible by 7 forms an AP:
105,112,119,…..,994
Where, a=105, d=7, a_n=994
Using:
a_n=a+(n-1)d
994=105+(n-1)7
994-105=(n-1)7
889=(n-1)7
\frac{889}{7}=n-1
127=n-1
n=127+1
n=128
Therefore, there are 128 three digit numbers divisible by 7.
*4. How many multiples of 4 lie between 10 and 250?
Solution:
Smallest multiple of 4 greater than 10 = 12
Largest multiple of 4 less than 250 = 248
So the sequence is:
12,16,20,…..248
This is an AP with:
a=12, d=4, a_n=248
Using:
a_n=a+(n-1)d
248=12+(n-1)4
248-12=(n-1)4
236=(n-1)4
\frac{236}{4}=n-1
59=n-1
n=59+1
n=60
Therefore, 60 multiples of 4 lie between 10 and 250.
*5. Find a GP for which the sum of the first two terms is – 4 and the fifth term is 4 times the third term.
Solution:
Let the first term is a and common ratio is r
So the GP is:
a, ar, ar^2, ar^3, ar^4, …..
Given, Sum of first two terms = -4
⇒ a+ar=-4
⇒ a(1+r)=-4 …….. (1)
Also, fifth term is 4 times the third term
ar^4=4ar^2
\frac{ar^4}{ar^2}=4
r^2=4
r=±\sqrt{4}
r=±2
⇒ r=2 or r=-2
Case 1: r=2
From (1)
a(1+2)=-4
a(1+2)=-4
3a=-4
a=\frac{-4}{3}
So the GP is : \frac{-4}{3}, \frac{-8}{3}, \frac{-16}{3}, \frac{-32}{3},…..
Case 2: r=-2
From (1)
a(1-2)=-4
a(-1)=-4
a=4
So the GP is : 4, -8, 16, -32,….
*6. Find all possible ways of expressing 100 as the sum of consecutive natural numbers.
Solution:
Consecutive natural numbers forms an AP with common difference 1.
Given: S_n=100, d=1
Sum, S_n=\frac{n}{2}[2a+(n-1)d]
100=\frac{n}{2}[2a+(n-1)1]
100=\frac{n}{2}(2a+n-1)
200=n(2a+n-1)
Now find possible values of n.
When n=5
200=5(2a+5-1)
200=5(2a+4)
200=10a+20
10a=200-20
10a=180
a=\frac{180}{10}
a=18
Hence,
100=18+19+20+21+22
When n=8
200=8(2a+8-1)
200=8(2a+7)
\frac{200}{8}=2a+7
25=2a+7
2a=25-7
2a=18
a=9
Hence,
100=9+10+11+12+13+14+15+16
So all possible ways are:
100=18+19+20+21+22
100=9+10+11+12+13+14+15+16
*7. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the 2^{nd} hour, 4^{th} hour and n^{th} hour?
Solution:
The number of bacteria double every hour, so this is a GP with common ratio 2.
Given: a=30, r=2
The GP is:
30,60,120,240,….
Since 30 is the number of bacteria at 0 hours,
So number of bacteria after n hours:
a_n=ar^n
Bacteria at the end of 2^{nd} hour:
a_2=30×2^2
a_2=30×4
a_2=120
Bacteria at the end of 4^{th} hour:
a_4=30×2^4
a_4=30×16
a_4=480
Bacteria at the end of n^{th} hour:
a_n=30×2^n
*8. The sum of the 4^{th} and 8^{th} terms of an AP is 24 and the sum of the 6^{th} and 10^{th} terms is 44. Find the first three terms of the AP.
Solution:
Let the first term be a common difference d.
Given, a_4+a_8=24
a+3d+a+7d=24
2a+10d=24
Dividing by 2 bothsides
a+5d=12…..(1)
Also, a_6+a_{10}=44
a+5d+a+9d=44
2a+14d=44
Dividing by 2 bothsides
a+7d=22…..(2)
Subtract (1) from (2)
a+7d-a-5d=22-12
2d=10
d=5 Put in (1)
a+5(5)=12
a+25=12
a=12-25
a=-13
∴ first three terms are :
a, a+d, a+2d
-13, -8, -3
*9. Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.
Solution:
The natural numbers (1,2,3,4,…) form an AP.
Here, a=1, d=1
Sum of n terms of AP:
S_n=\frac{n}{2}[2a+(n-1)d]
S_n=\frac{n}{2}[2×1+(n-1)1]
S_n=\frac{n}{2}(2+n-1)
S_n=\frac{n(n-1)}{2}
Given:
\frac{n(n+1)}{2}>1000
∴ n(n+1)>2000
For n = 44:
S_{44}=\frac{44\times45}{2}=990
For n = 45:
S_{45}=\frac{45\times46}{2}=1035
Since 1035 > 1000, the required smallest natural number is:
n=45
*10. Which term of the GP: 2, 8, 32, … is 131072? Write the explicit formula as well as the recursive formula for the n^{th} term.
Solution:
Given GP: 2,8,32,…
Here, a = 2,
and r=\frac{8}{2}=4
The nth term of a GP is:
a_n=ar^{n-1}
Let a_n=131072
2(4)^{n-1}=131072
Since 4 = 2² and 131072 = 2¹⁷:
2(2^2)^{n-1}=2^{17}
2(2)^{2(n-1)}=2^{17}
2^{1+2(n-1)}=2^{17}
2^{1+2n-2}=2^{17}
2^{2n-1}=2^{17}
So:
2n-1=17
2n=18
n=9
Therefore, 131072 is the 9th term.
Explicit formula:
a_n=2(4)^{n-1}
Recursive formula:
a_1=2,\quad a_n=4a_{n-1}\quad(n\ge2)
*11. The sum of the first three terms of a GP is \frac{13}{12} and their product is –1. Find the common ratio and the terms.
Solution:
Let the first three terms of GP be:
a,\ ar,\ ar^2
Their product is:
a(ar)(ar^2)=-1
a^3r^3=-1
(ar)^3=-1
∴ ar=-1 ….(1)
a=-\frac{1}{r} ….(2)
The sum of three terms is:
a+ar+ar^2=\frac{13}{12}
a(1+r+r^2)=\frac{13}{12}
-\frac{1}{r}(1+r+r^2)=\frac{13}{12} …[from (2)]
12(1+r+r^2)=-13r
12+12r+12r^2=-13r
12+12r+12r^2+13r=0
12r^2+25r+12=0
12r^2+16r+9r+12=0
4r(3r+4)+3(3r+4)=0
(3r+4)(4r+3)=0
Thus:
r=-\frac{4}{3}\ \text{or}\ r=-\frac{3}{4}
Case 1: If r=−\frac{4}{3}, Put in (1)
a(-\frac{4}{3})=-1
a=\frac{3}{4}
The terms are:
\frac{3}{4},\ -1,\ \frac{4}{3}
Case 2: If r=−\frac{3}{4}, Put in (1)
a(-\frac{3}{4})=-1
a=\frac{4}{3}
The terms are:
\frac{4}{3},\ -1,\ \frac{3}{4}
*12. If the 4^{th}, 10^{th} and 16^{th} terms of a GP are x, y and z respectively, prove that x, y, z are in GP.
Solution:
Let the first term of the GP be a and the common ratio be r.
The n^{th} term is:
a_n=ar^{n-1}
Therefore:
x=a_4=ar^3 ….(1)
y=a_{10}=ar^9 ….(2)
z=a_{16}=ar^{15} ….(3)
Squaring eq(2)
y^2=(ar^9)^2=a^2r^{18} ….(4)
multipling eq(1) & eq(3)
xz=(ar^3)(ar^{15})=a^2r^{18} ….(5)
From eq(4) & eq(5)
y^2=xz
yy=xz
\frac{y}{z}=\frac{z}{y}
Hence, x, y and z are in GP.
*13. The sum of the first three terms of a geometric progression is 26, and the sum of their squares is 364. Find the terms of the GP.
Solution:
Let the first three terms of GP be:
a,\ ar,\ ar^2
Given:
a+ar+ar^2=26 …(1)
Also:
a^2+a^2r^2+a^2r^4=364 …(2)
Squaring eq(1) bothsides
(a+ar+ar^2)^2=26^2 …(3)
Using (x + y + z)² = x² + y² + z² + 2(xy + yz + zx):
(a+ar+ar^2)^2=a^2+a^2r^2+a^2r^4+2(a\cdot ar+ar\cdot ar^2+ar^2\cdot a)
(26)^2=364+2(a\cdot ar+ar\cdot ar^2+ar^2\cdot a) [from (2) & (3)]
676-364=2(a\cdot ar+ar\cdot ar^2+ar^2\cdot a)
312=2(a\cdot ar+ar\cdot ar^2+ar^2\cdot a)
\frac{312}{2}=(a\cdot ar+ar\cdot ar^2+ar^2\cdot a)
156=ar(a+ar+ar^2)
156=ar(26) [from (1)]
\frac{156}{26}=ar
ar=6
a=\frac{6}{r} Put in eq(1)
a+ar+ar^2=26
\frac{6}{r}+\frac{6}{r}\cdot r+\frac{6}{r}\cdot r^2=26
\frac{6}{r}+6+6r=26
\frac{6}{r}+6r=26-6
\frac{6}{r}+6r=20
Multiplying by r:
6+6r^2=20r
2(3+3r^2)=2(10r)
3+3r^2=10r
3r^2-10r+3=0
Factorising:
(r-3)(3r-1)=0
r-3=0\ \text{or}\ 3r-1=0
r=3\ \text{or}\ r=\frac{1}{3}
Case 1: If r = 3, then a = 2
∴ The first three terms are:
2,\ 6,\ 18
Case 2: If r =\frac{1}{3}, then a = 18.
∴ The first three terms are:
18,\ 6,\ 2
*14. Suppose P_1=1, \ P_2 and for n>2, \ P_n=P_1+P_2+…+P_{n-1}+1. Find the values of P_1, \ P_2, \ …, \ P_8. Can you find a simpler recursive formula for P_n? Can you give an explicit formula?
Solution:
P_1=1
P_2=2
P_n=P_1+P_2+…+P_{n-1}+1, \quad n>2
For n = 3:
P_3=P_1+P_2+1=1+2+1=4
For n = 4:
P_4=P_1+P_2+P_3+1=1+2+4+1=8
For n=5
P_5=P_1+P_2+P_3+P_4+1=1+2+4+8+1=16
For n=6
P_6=P_1+P_2+P_3+P_4+P_5+1=1+2+4+8+16+1=32
For n=7
P_7=P_1+P_2+P_3+P_4+P_5+P_6+1=1+2+4+8+16+32+1=64
For n=8
P_8=P_1+P_2+P_3+P_4+P_5+P_6+P_7+1=1+2+4+8+16+32+64+1=128
Therefore, the sequence is:
1,\ 2,\ 4,\ 8,\ 16,\ 32,\ 64,\ 128 \ldots
Every term is twice the previous term.
∴ Simple recursive formula is:
P_1=1,\quad P_n=2P_{n-1}\quad(n\ge2)
Explicit formula:
P_n=2^{n-1}
*15. Suppose W_1,\ W_2, and for n>2, \ W_n=W_1+W_2+…+W_{n-2}+2. Find the values of W_1, \ W_2, \ …, \ W_8[/latex]. Do you recognise this sequence?
Solution:
W_1=1
W_2=2
W_n=W_1+W_2+…+W_{n-2}+2 \quad n>2
For n = 3:
W_3=W_1+2=1+2=3
For n = 4:
W_4=W_1+W_2+2=1+2+2=5
For n = 5:
W_5=W_1+W_2+W_3+2=1+2+3+2=8
For n = 6:
W_6=W_1+W_2+W_3+W_4+2=1+2+3+5+2=13
For n = 7:
W_7=W_1+W_2+W_3+W_4+W_5+2=1+2+3+5+8+2=21
For n = 8:
W_8=W_1+W_2+W_3+W_4+W_5+W_6+2=1+2+3+5+8+13+2=34
Therefore, the sequence is:
1,\ 2,\ 3,\ 5,\ 8,\ 13,\ 21,\ 34,\ \ldots
Each term is the sum of the previous two terms.
∴ it is Virahanka sequence (fibonacci sequence).
Simple recursive formula:
W_1=1,\quad W_2=2
W_n=W_{n-1}+W_{n-2}\quad(n\ge3)