RD Sharma Class 11 Solutions Chapter - Arithmetic Progressions | Mathematics (Maths) Class 11 - Commerce PDF Download

 Page 1


19. Arithmetic Progressions
Exercise 19.1
1. Question
If the n
th
 term of a sequence is given by a
n
 = n
2
 – n+1, write down its first five terms.
Answer
Given,
a
n
 = n
2
 – n+1
We can find first five terms of this sequence by putting values of n from 1 to 5.
When n = 1:
a
1
 = (1)
2
 – 1 + 1
? a
1
 = 1 – 1 + 1
? a
1
 = 1
When n = 2:
a
2
 = (2)
2
 – 2 + 1
? a
2
 = 4 – 2 + 1
? a
2
 = 3
When n = 3:
a
3
 = (3)
2
 – 3 + 1
? a
3
 = 9 – 3 + 1
? a
3
 = 7
When n = 4:
a
4
 = (4)
2
 – 4 + 1
? a
4
 = 16 – 4 + 1
? a
4
 = 13
When n = 5:
a
5
 = (5)
2
 – 5 + 1
? a
5
 = 25 – 5 + 1
? a
5
 = 21
? First five terms of the sequence are 1, 3, 7, 13, 21.
2. Question
A sequence is defined by a
n
 = n
3
 – 6n
2
 + 11n – 6, n ? N. Show that the first three terms of the sequence are
zero and all other terms are positive.
Answer
Given,
a
n
 = n
3
 – 6n
2
 + 11n – 6, n ? N
Page 2


19. Arithmetic Progressions
Exercise 19.1
1. Question
If the n
th
 term of a sequence is given by a
n
 = n
2
 – n+1, write down its first five terms.
Answer
Given,
a
n
 = n
2
 – n+1
We can find first five terms of this sequence by putting values of n from 1 to 5.
When n = 1:
a
1
 = (1)
2
 – 1 + 1
? a
1
 = 1 – 1 + 1
? a
1
 = 1
When n = 2:
a
2
 = (2)
2
 – 2 + 1
? a
2
 = 4 – 2 + 1
? a
2
 = 3
When n = 3:
a
3
 = (3)
2
 – 3 + 1
? a
3
 = 9 – 3 + 1
? a
3
 = 7
When n = 4:
a
4
 = (4)
2
 – 4 + 1
? a
4
 = 16 – 4 + 1
? a
4
 = 13
When n = 5:
a
5
 = (5)
2
 – 5 + 1
? a
5
 = 25 – 5 + 1
? a
5
 = 21
? First five terms of the sequence are 1, 3, 7, 13, 21.
2. Question
A sequence is defined by a
n
 = n
3
 – 6n
2
 + 11n – 6, n ? N. Show that the first three terms of the sequence are
zero and all other terms are positive.
Answer
Given,
a
n
 = n
3
 – 6n
2
 + 11n – 6, n ? N
We can find first three terms of sequence by putting the values of n form 1 to 3.
When n = 1:
a
1
 = (1)
3
 – 6(1)
2
 + 11(1) – 6
? a
1
 = 1 – 6 + 11 – 6
? a
1
 = 12 – 12
? a
1
 = 0
When n = 2:
a
2
 = (2)
3
 – 6(2)
2
 + 11(2) – 6
? a
2
 = 8 – 6(4) + 22 – 6
? a
2
 = 8 – 24 + 22 – 6
? a
2
 = 30 – 30
? a
2
 = 0
When n = 3:
a
3
 = (3)
3
 – 6(3)
2
 + 11(3) – 6
? a
3
 = 27 – 6(9) + 33 – 6
? a
3
 = 27 – 54 + 33 – 6
? a
3
 = 60 – 60
? a
3
 = 0
This shows that the first three terms of the sequence is zero.
When n = n:
a
n
 = n
3
 – 6n
2
 + 11n – 6
? a
n
 = n
3
 – 6n
2
 + 11n – 6 – n + n – 2 + 2
? a
n
 = n
3
 – 6n
2
 + 12n – 8 – n + 2
? a
n
 = (n)
3
 – 3×2n(n – 2) – (2)
3
 – n + 2
{(a – b)
3
 = (a)
3
 – (b)
3
 – 3ab(a – b)}
? a
n
 = (n – 2)
3
 – (n – 2)
Here, n – 2 will always be positive for n > 3
? a
n
 is always positive for n > 3
3. Question
Find the first four terms of the sequence defined by a
1
 = 3 and a
n
 = 3a
n–1
 + 2, for all n > 1.
Answer
Given,
a
1
 = 3 and a
n
 = 3a
n–1
 + 2, for all n > 1
We can find the first four terms of a sequence by putting values of n
from 1 to 4
Page 3


19. Arithmetic Progressions
Exercise 19.1
1. Question
If the n
th
 term of a sequence is given by a
n
 = n
2
 – n+1, write down its first five terms.
Answer
Given,
a
n
 = n
2
 – n+1
We can find first five terms of this sequence by putting values of n from 1 to 5.
When n = 1:
a
1
 = (1)
2
 – 1 + 1
? a
1
 = 1 – 1 + 1
? a
1
 = 1
When n = 2:
a
2
 = (2)
2
 – 2 + 1
? a
2
 = 4 – 2 + 1
? a
2
 = 3
When n = 3:
a
3
 = (3)
2
 – 3 + 1
? a
3
 = 9 – 3 + 1
? a
3
 = 7
When n = 4:
a
4
 = (4)
2
 – 4 + 1
? a
4
 = 16 – 4 + 1
? a
4
 = 13
When n = 5:
a
5
 = (5)
2
 – 5 + 1
? a
5
 = 25 – 5 + 1
? a
5
 = 21
? First five terms of the sequence are 1, 3, 7, 13, 21.
2. Question
A sequence is defined by a
n
 = n
3
 – 6n
2
 + 11n – 6, n ? N. Show that the first three terms of the sequence are
zero and all other terms are positive.
Answer
Given,
a
n
 = n
3
 – 6n
2
 + 11n – 6, n ? N
We can find first three terms of sequence by putting the values of n form 1 to 3.
When n = 1:
a
1
 = (1)
3
 – 6(1)
2
 + 11(1) – 6
? a
1
 = 1 – 6 + 11 – 6
? a
1
 = 12 – 12
? a
1
 = 0
When n = 2:
a
2
 = (2)
3
 – 6(2)
2
 + 11(2) – 6
? a
2
 = 8 – 6(4) + 22 – 6
? a
2
 = 8 – 24 + 22 – 6
? a
2
 = 30 – 30
? a
2
 = 0
When n = 3:
a
3
 = (3)
3
 – 6(3)
2
 + 11(3) – 6
? a
3
 = 27 – 6(9) + 33 – 6
? a
3
 = 27 – 54 + 33 – 6
? a
3
 = 60 – 60
? a
3
 = 0
This shows that the first three terms of the sequence is zero.
When n = n:
a
n
 = n
3
 – 6n
2
 + 11n – 6
? a
n
 = n
3
 – 6n
2
 + 11n – 6 – n + n – 2 + 2
? a
n
 = n
3
 – 6n
2
 + 12n – 8 – n + 2
? a
n
 = (n)
3
 – 3×2n(n – 2) – (2)
3
 – n + 2
{(a – b)
3
 = (a)
3
 – (b)
3
 – 3ab(a – b)}
? a
n
 = (n – 2)
3
 – (n – 2)
Here, n – 2 will always be positive for n > 3
? a
n
 is always positive for n > 3
3. Question
Find the first four terms of the sequence defined by a
1
 = 3 and a
n
 = 3a
n–1
 + 2, for all n > 1.
Answer
Given,
a
1
 = 3 and a
n
 = 3a
n–1
 + 2, for all n > 1
We can find the first four terms of a sequence by putting values of n
from 1 to 4
When n = 1:
a
1
 = 3
When n = 2:
a
2
 = 3a
2–1
 + 2
? a
2
 = 3a
1
 + 2
? a
2
 = 3(3) + 2
? a
2
 = 9 + 2
? a
2
 = 11
When n = 3:
a
3
 = 3a
3–1
 + 2
? a
3
 = 3a
2
 + 2
? a
3
 = 3(11) + 2
? a
3
 = 33 + 2
? a
3
 = 35
When n = 4:
a
4
 = 3a
4–1
 + 2
? a
4
 = 3a
3
 + 2
? a
4
 = 3(35) + 2
? a
4
 = 105 + 2
? a
4
 = 107
? First four terms of sequence are 3, 11, 35, 107.
4 A. Question
Write the first five terms in each of the following sequences:
a
1
 = 1, a
n
 = a
n–1
 + 2, n > 1
Answer
Given,
a
1
 = 1, a
n
 = a
n–1
 + 2, n > 1
We can find the first five terms of a sequence by putting values of n
from 1 to 5
When n = 1:
a
1
 = 1
When n = 2:
a
2
 = a
2–1
 + 2
? a
2
 = a
1
 + 2
? a
2
 = 1 + 2
Page 4


19. Arithmetic Progressions
Exercise 19.1
1. Question
If the n
th
 term of a sequence is given by a
n
 = n
2
 – n+1, write down its first five terms.
Answer
Given,
a
n
 = n
2
 – n+1
We can find first five terms of this sequence by putting values of n from 1 to 5.
When n = 1:
a
1
 = (1)
2
 – 1 + 1
? a
1
 = 1 – 1 + 1
? a
1
 = 1
When n = 2:
a
2
 = (2)
2
 – 2 + 1
? a
2
 = 4 – 2 + 1
? a
2
 = 3
When n = 3:
a
3
 = (3)
2
 – 3 + 1
? a
3
 = 9 – 3 + 1
? a
3
 = 7
When n = 4:
a
4
 = (4)
2
 – 4 + 1
? a
4
 = 16 – 4 + 1
? a
4
 = 13
When n = 5:
a
5
 = (5)
2
 – 5 + 1
? a
5
 = 25 – 5 + 1
? a
5
 = 21
? First five terms of the sequence are 1, 3, 7, 13, 21.
2. Question
A sequence is defined by a
n
 = n
3
 – 6n
2
 + 11n – 6, n ? N. Show that the first three terms of the sequence are
zero and all other terms are positive.
Answer
Given,
a
n
 = n
3
 – 6n
2
 + 11n – 6, n ? N
We can find first three terms of sequence by putting the values of n form 1 to 3.
When n = 1:
a
1
 = (1)
3
 – 6(1)
2
 + 11(1) – 6
? a
1
 = 1 – 6 + 11 – 6
? a
1
 = 12 – 12
? a
1
 = 0
When n = 2:
a
2
 = (2)
3
 – 6(2)
2
 + 11(2) – 6
? a
2
 = 8 – 6(4) + 22 – 6
? a
2
 = 8 – 24 + 22 – 6
? a
2
 = 30 – 30
? a
2
 = 0
When n = 3:
a
3
 = (3)
3
 – 6(3)
2
 + 11(3) – 6
? a
3
 = 27 – 6(9) + 33 – 6
? a
3
 = 27 – 54 + 33 – 6
? a
3
 = 60 – 60
? a
3
 = 0
This shows that the first three terms of the sequence is zero.
When n = n:
a
n
 = n
3
 – 6n
2
 + 11n – 6
? a
n
 = n
3
 – 6n
2
 + 11n – 6 – n + n – 2 + 2
? a
n
 = n
3
 – 6n
2
 + 12n – 8 – n + 2
? a
n
 = (n)
3
 – 3×2n(n – 2) – (2)
3
 – n + 2
{(a – b)
3
 = (a)
3
 – (b)
3
 – 3ab(a – b)}
? a
n
 = (n – 2)
3
 – (n – 2)
Here, n – 2 will always be positive for n > 3
? a
n
 is always positive for n > 3
3. Question
Find the first four terms of the sequence defined by a
1
 = 3 and a
n
 = 3a
n–1
 + 2, for all n > 1.
Answer
Given,
a
1
 = 3 and a
n
 = 3a
n–1
 + 2, for all n > 1
We can find the first four terms of a sequence by putting values of n
from 1 to 4
When n = 1:
a
1
 = 3
When n = 2:
a
2
 = 3a
2–1
 + 2
? a
2
 = 3a
1
 + 2
? a
2
 = 3(3) + 2
? a
2
 = 9 + 2
? a
2
 = 11
When n = 3:
a
3
 = 3a
3–1
 + 2
? a
3
 = 3a
2
 + 2
? a
3
 = 3(11) + 2
? a
3
 = 33 + 2
? a
3
 = 35
When n = 4:
a
4
 = 3a
4–1
 + 2
? a
4
 = 3a
3
 + 2
? a
4
 = 3(35) + 2
? a
4
 = 105 + 2
? a
4
 = 107
? First four terms of sequence are 3, 11, 35, 107.
4 A. Question
Write the first five terms in each of the following sequences:
a
1
 = 1, a
n
 = a
n–1
 + 2, n > 1
Answer
Given,
a
1
 = 1, a
n
 = a
n–1
 + 2, n > 1
We can find the first five terms of a sequence by putting values of n
from 1 to 5
When n = 1:
a
1
 = 1
When n = 2:
a
2
 = a
2–1
 + 2
? a
2
 = a
1
 + 2
? a
2
 = 1 + 2
? a
2
 = 3
When n = 3:
a
3
 = a
3–1
 + 2
? a
3
 = a
2
 + 2
? a
3
 = 3 + 2
? a
3
 = 5
When n = 4:
a
4
 = a
4–1
 + 2
? a
4
 = a
3
 + 2
? a
4
 = 5 + 2
? a
4
 = 7
When n = 5:
a
5
 = a
5–1
 + 2
? a
5
 = a
4
 + 2
? a
5
 = 7 + 2
? a
5
 = 9
? First five terms of the sequence are 1, 3, 5, 7, 9.
4 B. Question
Write the first five terms in each of the following sequences:
a
1
 = 1 = a
2
, a
n
 = a
n–1
 + a
n–2
, n>2
Answer
Given,
a
1
 = 1 = a
2
, a
n
 = a
n–1
 + a
n–2
, n>2
We can find the first five terms of a sequence by putting values of n
from 1 to 5
When n = 1:
a
1
 = 1
When n = 2:
a
1
 = 1
When n = 3:
a
3
 = a
3–1
 + a
3–2
? a
3
 = a
2
 + a
1
? a
3
 = 1 + 1
? a
3
 = 2
When n = 4:
Page 5


19. Arithmetic Progressions
Exercise 19.1
1. Question
If the n
th
 term of a sequence is given by a
n
 = n
2
 – n+1, write down its first five terms.
Answer
Given,
a
n
 = n
2
 – n+1
We can find first five terms of this sequence by putting values of n from 1 to 5.
When n = 1:
a
1
 = (1)
2
 – 1 + 1
? a
1
 = 1 – 1 + 1
? a
1
 = 1
When n = 2:
a
2
 = (2)
2
 – 2 + 1
? a
2
 = 4 – 2 + 1
? a
2
 = 3
When n = 3:
a
3
 = (3)
2
 – 3 + 1
? a
3
 = 9 – 3 + 1
? a
3
 = 7
When n = 4:
a
4
 = (4)
2
 – 4 + 1
? a
4
 = 16 – 4 + 1
? a
4
 = 13
When n = 5:
a
5
 = (5)
2
 – 5 + 1
? a
5
 = 25 – 5 + 1
? a
5
 = 21
? First five terms of the sequence are 1, 3, 7, 13, 21.
2. Question
A sequence is defined by a
n
 = n
3
 – 6n
2
 + 11n – 6, n ? N. Show that the first three terms of the sequence are
zero and all other terms are positive.
Answer
Given,
a
n
 = n
3
 – 6n
2
 + 11n – 6, n ? N
We can find first three terms of sequence by putting the values of n form 1 to 3.
When n = 1:
a
1
 = (1)
3
 – 6(1)
2
 + 11(1) – 6
? a
1
 = 1 – 6 + 11 – 6
? a
1
 = 12 – 12
? a
1
 = 0
When n = 2:
a
2
 = (2)
3
 – 6(2)
2
 + 11(2) – 6
? a
2
 = 8 – 6(4) + 22 – 6
? a
2
 = 8 – 24 + 22 – 6
? a
2
 = 30 – 30
? a
2
 = 0
When n = 3:
a
3
 = (3)
3
 – 6(3)
2
 + 11(3) – 6
? a
3
 = 27 – 6(9) + 33 – 6
? a
3
 = 27 – 54 + 33 – 6
? a
3
 = 60 – 60
? a
3
 = 0
This shows that the first three terms of the sequence is zero.
When n = n:
a
n
 = n
3
 – 6n
2
 + 11n – 6
? a
n
 = n
3
 – 6n
2
 + 11n – 6 – n + n – 2 + 2
? a
n
 = n
3
 – 6n
2
 + 12n – 8 – n + 2
? a
n
 = (n)
3
 – 3×2n(n – 2) – (2)
3
 – n + 2
{(a – b)
3
 = (a)
3
 – (b)
3
 – 3ab(a – b)}
? a
n
 = (n – 2)
3
 – (n – 2)
Here, n – 2 will always be positive for n > 3
? a
n
 is always positive for n > 3
3. Question
Find the first four terms of the sequence defined by a
1
 = 3 and a
n
 = 3a
n–1
 + 2, for all n > 1.
Answer
Given,
a
1
 = 3 and a
n
 = 3a
n–1
 + 2, for all n > 1
We can find the first four terms of a sequence by putting values of n
from 1 to 4
When n = 1:
a
1
 = 3
When n = 2:
a
2
 = 3a
2–1
 + 2
? a
2
 = 3a
1
 + 2
? a
2
 = 3(3) + 2
? a
2
 = 9 + 2
? a
2
 = 11
When n = 3:
a
3
 = 3a
3–1
 + 2
? a
3
 = 3a
2
 + 2
? a
3
 = 3(11) + 2
? a
3
 = 33 + 2
? a
3
 = 35
When n = 4:
a
4
 = 3a
4–1
 + 2
? a
4
 = 3a
3
 + 2
? a
4
 = 3(35) + 2
? a
4
 = 105 + 2
? a
4
 = 107
? First four terms of sequence are 3, 11, 35, 107.
4 A. Question
Write the first five terms in each of the following sequences:
a
1
 = 1, a
n
 = a
n–1
 + 2, n > 1
Answer
Given,
a
1
 = 1, a
n
 = a
n–1
 + 2, n > 1
We can find the first five terms of a sequence by putting values of n
from 1 to 5
When n = 1:
a
1
 = 1
When n = 2:
a
2
 = a
2–1
 + 2
? a
2
 = a
1
 + 2
? a
2
 = 1 + 2
? a
2
 = 3
When n = 3:
a
3
 = a
3–1
 + 2
? a
3
 = a
2
 + 2
? a
3
 = 3 + 2
? a
3
 = 5
When n = 4:
a
4
 = a
4–1
 + 2
? a
4
 = a
3
 + 2
? a
4
 = 5 + 2
? a
4
 = 7
When n = 5:
a
5
 = a
5–1
 + 2
? a
5
 = a
4
 + 2
? a
5
 = 7 + 2
? a
5
 = 9
? First five terms of the sequence are 1, 3, 5, 7, 9.
4 B. Question
Write the first five terms in each of the following sequences:
a
1
 = 1 = a
2
, a
n
 = a
n–1
 + a
n–2
, n>2
Answer
Given,
a
1
 = 1 = a
2
, a
n
 = a
n–1
 + a
n–2
, n>2
We can find the first five terms of a sequence by putting values of n
from 1 to 5
When n = 1:
a
1
 = 1
When n = 2:
a
1
 = 1
When n = 3:
a
3
 = a
3–1
 + a
3–2
? a
3
 = a
2
 + a
1
? a
3
 = 1 + 1
? a
3
 = 2
When n = 4:
a
4
 = a
4–1
 + a
4–2
? a
4
 = a
3
 + a
2
? a
4
 = 2 + 1
? a
4
 = 3
When n = 5:
a
5
 = a
5–1
 + a
5–2
? a
5
 = a
4
 + a
3
? a
5
 = 3 + 2
? a
5
 = 5
? First five terms of the sequence are 1, 1, 2, 3, 5.
4 C. Question
Write the first five terms in each of the following sequences:
a
1
 = a
2
 =2, a
n
 = a
n–1
 – 1, n>2
Answer
Given,
a
1
 = 2 = a
2
, a
n
 = a
n–1
 – 1, n>2
We can find the first five terms of a sequence by putting values of n
from 1 to 5
When n = 1:
a
1
 = 2
When n = 2:
a
1
 = 2
When n = 3:
a
3
 = a
3–1
 – 1
? a
3
 = a
2
 – 1
? a
3
 = 2 – 1
? a
3
 = 1
When n = 4:
a
4
 = a
4–1
 – 1
? a
4
 = a
3
 – 1
? a
4
 = 1 – 1
? a
4
 = 0
When n = 5:
a
5
 = a
5–1
 – 1
? a
5
 = a
4
 – 1