Arithmetic progression

EXERCISE 19.4

QUESTION 1

Find the sum of the following arithmetic progressions:

(i) 50,46,42, \ldots to 10 terms

Sol :

We have :

a = 50

d = (46 -50) = -4

n = 10

S_{n}=\frac{n}{2}[2 a+(n-1) d]

=\frac{10}{2}[2 \times 50+(10-1)(-4)]

=5[100-36]

= 320

 

(ii) 1,3,5,7, \ldots to 12 terms

Sol :

We have :

a = 1

d = (3 – 1) = 2

n = 12

S_{n}=\frac{n}{2}[2 a+(n-1) d]

=\frac{12}{2}[2 \times 1+(12-1)(2)]

=6[24]

= 144

 

(iii) 3,9 / 2,6,15 / 2, \ldots to 25 terms

Sol :

We have :

a = 3

d =\left (\dfrac{9}{2}-3\right) = \dfrac{3}{2}

n = 25

S_{n}=\frac{n}{2}[2 a+(n-1) d]

=\frac{25}{2}[2 \times 3+(25-1)(\dfrac{3}{2})]

=\frac{25}{2} \times 42

= 525

 

(iv) 41,36,31, \ldots to 12 terms

Sol :

We have :

a = 41

d = (36 – 41) = -5

n = 12

S_{n}=\frac{n}{2}[2 a+(n-1) d]

=6 \times 27

= 162

 

(v) a+b, a-b, a-3 b, \ldots to 22 terms

Sol :

We have :

First term = a + b

d = (a – b – a – b) = -2b

n = 22

S_{n}=\frac{n}{2}[2 a+(n-1) d]

=\frac{22}{2}[2 \times(a+b)+(22-1)(-2 b)]

=11[2 a-40 b]

=22 a-440 b

 

(vi) (x-y)^{2},\left(x^{2}+y^{2}\right),(x+y)^{2}, \ldots to n terms

Sol :

We have :

a=(\mathrm{x}-\mathrm{y})^{2}

d=\left(x^{2}+y^{2}-(\mathrm{x}-\mathrm{y})^{2}\right)

=2 x y

S_{n}=\frac{n}{2}[2 a+(n-1) d]

=\frac{n}{2}\left[2(\mathrm{x}-\mathrm{y})^{2}+(n-1)(2 x y)\right]

=\frac{n}{2} \times 2\left[(\mathrm{x}-\mathrm{y})^{2}+(n-1)(x y)\right]

=n\left[(\mathrm{x}-\mathrm{y})^{2}+(n-1)(x y)\right]

 

(vii) \frac{x-y}{x+y}, \frac{3 x-2 y}{x+y}, \frac{5 x-3 y}{x+y}, \ldots to n terms

Sol :

We have :

a=\frac{x-y}{x+y}

d=\left(\frac{3 x-2 y}{x+y}-\frac{x-y}{x+y}\right)

=\left(\frac{2 x-y}{x+y}\right)

 

S_{n}=\frac{n}{2}[2 a+(n-1) d]

=\frac{n}{2}\left[2\left(\frac{x-y}{x+y}\right)+(n-1)\left(\frac{2 x-y}{x+y}\right)\right]

=\frac{n}{2(x+y)}[(2 x-2 y)+(2 x-y)(n-1)]

=\frac{n}{2(x+y)}[2 x-2 y-2 x+y+n(2 x-y)]

=\frac{n}{2(x+y)}[n(2 x-y)-y]

 

QUESTION 2

Find the sum of the following series:

(i) 2+5+8+\ldots+182

Sol :

Here, the series is an A.P. where we have the following:

a = 2

d = (5 – 2) = 3

a_{n}=182

\Rightarrow a+(n-1)(d)=a_{n}

\Rightarrow 2+(n-1)(3)=182

\Rightarrow 3 n-1=182

\Rightarrow 3 n=183

\Rightarrow n=61

 

S_{n}=\frac{n}{2}[2 a+(n-1) d]

\Rightarrow S_{61}=\frac{61}{2}[2 \times 2+(61-1) \times 3]

=\frac{61}{2}[2 \times 2+60 \times 3]

= 5412

 

(ii) 101+99+97+\ldots+47

Sol :

Here, the series is an A.P. where we have the following:

a = 101

d = (99 – 101) = -2

a_{n}=47

 

\Rightarrow a+(n-1)(d)=a_{n}

\Rightarrow 101+(n-1)(-2)=47

\Rightarrow 2 n-2=54

\Rightarrow 2 n=56

\Rightarrow n=28

 

S_{n}=\frac{n}{2}[2 a+(n-1) d]

\Rightarrow S_{28}=\frac{28}{2}[2 \times 101+(28-1) \times(-2)]

=\frac{28}{2}[2 \times 101+27 \times(-2)]

= 2072

 

(iii) (a-b)^{2}+\left(a^{2}+b^{2}\right)+(a+b)^{2}+\ldots+\left[(a+b)^{2}+6 a b\right]

Sol :

Here, the series is an A.P. where we have the following:

a = (a-b)^2

d=\left(a^{2}+b^{2}-(a-b)^{2}\right)

= 2ab

a_{n}=\left[(\mathrm{a}+\mathrm{b})^{2}+6 a b\right]

 

\Rightarrow a+(n-1)(d)=a_{n}

\Rightarrow(a-b)^{2}+(n-1)(2 a b)=\left[(a+b)^{2}+6 a b\right]

\Rightarrow a^{2}+b^{2}-2 a b+2 a b n-2 a b=\left[a^{2}+b^{2}+2 a b+6 a b\right]

\Rightarrow a^{2}+b^{2}-4 a b+2 a b n=a^{2}+b^{2}+8 a b

\Rightarrow 2 a b n=12 a b

\Rightarrow n=6

 

S_{n}=\frac{n}{2}[2 a+(n-1) d]

=3\left[2\left(a^{2}+b^{2}-2 a b\right)+10 a b\right]

=3\left[2 a^{2}+2 b^{2}-4 a b+10 a b\right]

=3\left[2 a^{2}+2 b^{2}+6 a b\right]

=3\left[2 a^{2}+2 b^{2}+6 a b\right]

=6\left[a^{2}+b^{2}+3 a b\right]

 

QUESTION 3

Find the sum of first n natural numbers.

Sol :

The first n natural numbers are: 1,2,3,4 \ldots

a = 1 , d = 1

Total terms = n

 

S_{n}=\frac{n}{2}[2 a+(n-1) d]

\Rightarrow S_{n}=\frac{n}{2}[2 \times 1+(n-1) 1]

\Rightarrow S_{n}=\frac{n}{2}[2+(n-1) 1]

\Rightarrow S_{n}=\frac{n}{2}[n+1]

 

QUESTION 4

Find the sum of all natural numbers between 1 and 100, which are divisible by 2 or 5 

Sol :

We have to find the sum of all the natural numbers from 1 to 100 , that are divisible by 2 or 5 .

S = A + B – C

S = Requires sum

A = sum of natural numbers between 1 to 100 divisible by 2 

B = sum of all the natural number between 1 to 100 divisible by 5

C = sum of all natural number between 1 to 100 , which is divisible by 2 and 5 both (just to remove those number which is already present in A and B i.e 10 )

 

A =(2+4+6+8+\ldots+98)

where a = 2 , a_n=98

 

B (5+10+15+\ldots+95)

where a = 5 , a_n=98

 

C (10+20+30+\ldots+90)

where a = 10 , a_n=90

 

S = (2+4+6+8+\ldots+98) +(5+10+15+\ldots+95) -(10+20+30+\ldots+90)

=\frac{50}{2}(2+98) +\frac{20}{2}(5+95) -\frac{10}{2}(10+90) \left[S_n=\dfrac{n}{2} \left(2a+(n-1)d\right)\right]

= 2500 + 1000 -500

= 3000

 

QUESTION 5

Find the sum of first n odd natural numbers.

Sol :

The first n odd natural numbers are: 1 , 3 , 5 , 7 …..

a = 1 , d = 2 

Total terms = n

S_{n}=\frac{n}{2}[2 a+(n-1) d]

\Rightarrow S_{n}=\frac{n}{2}[2 \times 1+(n-1) 2]

\Rightarrow S_{n}=\frac{n}{2}[2+(n-1) 2]

\Rightarrow S_{n}=\frac{n}{2}[2 n]

\Rightarrow S_{n}=n^{2}

 

QUESTION 6

Find the sum of all odd numbers between 100 and 200 

Sol :

All the odd numbers between 100 and 200 are:

101 , 103 , …… 199

Here , we have

a = 101

d = 2

a_{n}=199

 

\Rightarrow a+(n-1) d=a_n

\Rightarrow 101+(n-1) \times 2=199

\Rightarrow 2 n-2=98

\Rightarrow 2 n=100

\Rightarrow n=50

 

S_{n}=\frac{n}{2}[2 a+(n-1) d]

\Rightarrow S_{50}=\frac{50}{2}[2 \times 101+(50-1) 2]

\Rightarrow S_{50}=25[202+98]

\Rightarrow S_{50}=7500

 

QUESTION 7

Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667

Sol :

The odd integers between 1 and 1000 that are divisible by 3 are:

3 , 9 , 15 , 21 ….. 999

Here , we have

a = 3 , d = 6 

a_{n}=999

 

\Rightarrow a+(n-1) d=a_n

\Rightarrow 3+(n-1) 6=999

\Rightarrow 6 n=1002

\Rightarrow n=167

 

S_{n}=\frac{n}{2}[2 a+(n-1) d]

\Rightarrow S_{167}=\frac{167}{2}[2 \times 3+(167-1) 6]

\Rightarrow S_{167}=\frac{167}{2}[1002]

=83667

Hence , proved

 

QUESTION 8

Find the sum of all integers between 84 and 719, which are multiples of 5

Sol :

The integers between 84 and 719, which are multiples of 5 are:

85 , 90 . . .715

Here , we have

a = 5 , d = 5 

a_{n}=715

 

\Rightarrow a+(n-1) d=a_n

\Rightarrow 85+(n-1) 5=715

\Rightarrow 5 n-5=630

\Rightarrow n=127

 

S_{n}=\frac{n}{2}[2 a+(n-1) d]

\Rightarrow S_{127}=\frac{127}{2}[2 \times 85+(127-1) 5]

\Rightarrow S_{127}=\frac{127}{2}[800]

=50800

 

QUESTION 9

 

QUESTION 10

QUESTION 11

QUESTION 12

Find the sum of the series :

3 + 5 + 7 + 6 + 9  + 13 + 17 + …… to 3terms .

Sol :

The given sequence can be rewritten as 3 + 6 + 9… to n terms + 5 + 9 + 13 +. … to terms + 7 + 12 + 17+ … to n terms

Clearly , all these sequence forms an A.P. having n  terms with first terms 3 , 5 , 7 and common difference 3 , 4 , 5

 

Hence , required sum  =\frac{n}{2}[2 \times 3+(n-1) 3] +\frac{n}{2}[2 \times 5+(n-1) 4] +\frac{n}{2}[2 \times 7+(n-1) 5]

=\frac{n}{2}[(6+3 n-3) +(10+4 n-4) +(14+5 n-5) ]

=\frac{n}{2}[12 n+18]

=3 n(2 n+3)

 

QUESTION 13

Find the sum of all those integers between 100 and 800 each of which on division by 16 leaves the
remainder 7

Sol :

The sum of all those integers between 100 and 800 each of which on division by 16 leaves the
remainder 7 are:

 

QUESTION 14

QUESTION 15

QUESTION 16

QUESTION 17

QUESTION 18

QUESTION 19

QUESTION 20

QUESTION 21

QUESTION 22

QUESTION 23

QUESTION 24

QUESTION 25

QUESTION 26

QUESTION 27

QUESTION 28

QUESTION 29

QUESTION 30

QUESTION 31

QUESTION 32

QUESTION 33

QUESTION 34

 

 

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