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RD Chapter 19- Arithmetic Progressions Ex-19.4 Interview Questions Answers

Question 1 :

Find the sum of the following arithmetic progressions:
(i) 50, 46, 42, …. to 10 terms

(ii) 1, 3, 5, 7, … to 12 terms

(iii) 3, 9/2, 6, 15/2, … to 25 terms

(iv) 41, 36, 31, … to 12 terms

(v) a+b, a-b, a-3b, … to 22 terms

(vi) (x – y)2, (x2 + y2), (x + y)2,… to n terms

(vii) (x – y)/(x + y), (3x – 2y)/(x + y), (5x – 3y)/(x + y), … to n terms

Answer 1 :

(i) 50, 46, 42, …. to 10terms

n = 10

First term, a = a1 =50

Common difference, d =a2 – a= 46 – 50 = -4

By using the formula,

S = n/2 (2a + (n – 1)d)

Substitute the valuesof ‘a’ and ‘d’, we get

S = 10/2 (100 + (9)(-4))

= 5 (100 – 36)

= 5 (64)

= 320

The sum of the givenAP is 320.

(ii) 1, 3, 5, 7, … to 12terms

n = 12

First term, a = a1 =1

Common difference, d =a2 – a= 3 – 1 = 2

By using the formula,

S = n/2 (2a + (n – 1)d)

Substitute the valuesof ‘a’ and ‘d’, we get

S = 12/2 (2(1) +(12-1) (2))

= 6 (2 + (11) (2))

= 6 (2 + 22)

= 6 (24)

= 144

The sum of the givenAP is 144.

(iii) 3, 9/2, 6, 15/2, … to25 terms

n = 25

First term, a = a1 =3

Common difference, d =a2 – a= 9/2 – 3 = (9 – 6)/2 = 3/2

By using the formula,

S = n/2 (2a + (n – 1)d)

Substitute the valuesof ‘a’ and ‘d’, we get

S = 25/2 (2(3) +(25-1) (3/2))

= 25/2 (6 + (24)(3/2))

= 25/2 (6 + 36)

= 25/2 (42)

= 25 (21)

= 525

The sum of the givenAP is 525.

(iv) 41, 36, 31, … to 12terms

n = 12

First term, a = a1 =41

Common difference, d =a2 – a= 36 – 41 = -5

By using the formula,

S = n/2 (2a + (n – 1)d)

Substitute the valuesof ‘a’ and ‘d’, we get

S = 12/2 (2(41) +(12-1) (-5))

= 6 (82 + (11) (-5))

= 6 (82 – 55)

= 6 (27)

= 162

The sum of the givenAP is 162.

(v) a+b, a-b, a-3b, … to22 terms

n = 22

First term, a = a1 =a+b

Common difference, d =a2 – a= (a-b) – (a+b) = a-b-a-b = -2b

By using the formula,

S = n/2 (2a + (n – 1)d)

Substitute the valuesof ‘a’ and ‘d’, we get

S = 22/2 (2(a+b) +(22-1) (-2b))

= 11 (2a + 2b + (21)(-2b))

= 11 (2a + 2b – 42b)

= 11 (2a – 40b)

= 22a – 440b

The sum of the givenAP is 22a – 440b.

(vi) (x – y)2,(x2 + y2), (x + y)2, … to n terms

n = n

First term, a = a1 =(x-y)2

Common difference, d =a2 – a= (x2 + y2)– (x-y)2 = 2xy

By using the formula,

S = n/2 (2a + (n – 1)d)

Substitute the valuesof ‘a’ and ‘d’, we get

S = n/2 (2(x-y)2 +(n-1) (2xy))

= n/2 (2 (x2 +y2 – 2xy) + 2xyn – 2xy)

= n/2 × 2 ((x2 +y2 – 2xy) + xyn – xy)

= n (x2 +y2 – 3xy + xyn)

The sum of the givenAP is n (x2 + y2 – 3xy + xyn).

(vii) (x – y)/(x + y), (3x –2y)/(x + y), (5x – 3y)/(x + y), … to n terms

n = n

First term, a = a1 =(x-y)/(x+y)

Common difference, d =a2 – a= (3x – 2y)/(x + y) – (x-y)/(x+y) = (2x– y)/(x+y)

By using the formula,

S = n/2 (2a + (n – 1)d)

Substitute the valuesof ‘a’ and ‘d’, we get

S = n/2(2((x-y)/(x+y)) + (n-1) ((2x – y)/(x+y)))

= n/2(x+y) {n (2x-y) –y}

The sum of the givenAP is n/2(x+y) {n (2x-y) – y}

Question 2 :

. Find the sum of the following series:
(i) 2 + 5 + 8 + … + 182

(ii) 101 + 99 + 97 + … + 47

(iii) (a – b)2 + (a2 + b2) +(a + b)2 + s…. + [(a + b)2 + 6ab]

Answer 2 :

(i) 2 + 5 + 8 + … + 182

First term, a = a1 =2

Common difference, d =a2 – a= 5 – 2 = 3

an termof given AP is 182

an = a+ (n-1) d

182 = 2 + (n-1) 3

182 = 2 + 3n – 3

182 = 3n – 1

3n = 182 + 1

n = 183/3

= 61

Now,

By using the formula,

S = n/2 (a + l)

= 61/2 (2 + 182)

= 61/2 (184)

= 61 (92)

= 5612

The sum of the seriesis 5612

(ii) 101 + 99 + 97 + … + 47

First term, a = a1 =101

Common difference, d =a2 – a= 99 – 101 = -2

an termof given AP is 47

an = a+ (n-1) d

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

47 = 101 – 2n + 2

2n = 103 – 47

2n = 56

n = 56/2 = 28

Then,

S = n/2 (a + l)

= 28/2 (101 + 47)

= 28/2 (148)

= 14 (148)

= 2072

The sum of the seriesis 2072

(iii) (a – b)2 +(a2 + b2) + (a + b)2 + s…. + [(a +b)2 + 6ab]

First term, a = a1 =(a-b)2

Common difference, d =a2 – a= (a2 + b2)– (a – b)2 = 2ab

an termof given AP is [(a + b)2 + 6ab]

an = a+ (n-1) d

[(a +b)2 + 6ab] = (a-b)2 + (n-1)2ab

a2 + b2 +2ab + 6ab = a2 + b2 – 2ab + 2abn – 2ab

a2 + b2 +8ab – a2 – b2 + 2ab + 2ab = 2abn

12ab = 2abn

n = 12ab / 2ab

= 6

Then,

S = n/2 (a + l)

= 6/2 ((a-b)2 +[(a + b)2 + 6ab])

= 3 (a2 +b2 – 2ab + a2 + b2 + 2ab + 6ab)

= 3 (2a2 +2b2 + 6ab)

= 3 × 2 (a2 +b2 + 3ab)

= 6 (a2 +b2 + 3ab)

The sum of the seriesis 6 (a2 + b2 + 3ab)

Question 3 :

Find the sum of first n natural numbers.

Answer 3 :

Let AP be 1, 2, 3, 4,…, n

Here,

First term, a = a1 =1

Common difference, d =a2 – a= 2 – 1 = 1

l = n

So, the sum of n terms= S = n/2 [2a + (n-1) d]

= n/2 [2(1) + (n-1) 1]

= n/2 [2 + n – 1]

= n/2 [n – 1]

The sum of the firstn natural numbers is n(n-1)/2

Question 4 :

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

Answer 4 :

The natural numberswhich are divisible by 2 or 5 are:

2 + 4 + 5 + 6 + 8 + 10+ … + 100 = (2 + 4 + 6 +…+ 100) + (5 + 15 + 25 +…+95)

Now, (2 + 4 + 6 +…+100) + (5 + 15 + 25 +…+95) are AP with common difference of 2 and 10.

So, for the 1st sequence=> (2 + 4 + 6 +…+ 100)

a = 2, d = 4-2 = 2, an =100

By using the formula,

an = a+ (n-1)d

100 = 2 + (n-1)2

100 = 2 + 2n – 2

2n = 100

n = 100/2

= 50

So now, S = n/2 (2a +(n-1)d)

= 50/2 (2(2) +(50-1)2)

= 25 (4 + 49(2))

= 25 (4 + 98)

= 2550

Again, for the 2nd sequence,(5 + 15 + 25 +…+95)

a = 5, d = 15-5 = 10,an = 95

By using the formula,

an = a+ (n-1)d

95 = 5 + (n-1)10

95 = 5 + 10n – 10

10n = 95 +10 – 5

10n = 100

n = 100/10

= 10

So now, S = n/2 (2a +(n-1)d)

= 10/2 (2(5) +(10-1)10)

= 5 (10 + 9(10))

= 5 (10 + 90)

= 500

The sum of thenumbers divisible by 2 or 5 is: 2550 + 500 = 3050

Question 5 :

Find the sum of first n odd natural numbers.

Answer 5 :

Given an AP of first nodd natural numbers whose first term a is 1, and common difference d is 3

The sequence is 1, 3,5, 7……n

a = 1, d = 3-1 = 2, n= n

By using the formula,

S = n/2 [2a + (n-1)d]

= n/2 [2(1) + (n-1)2]

= n/2 [2 + 2n – 2]

= n/2 [2n]

= n2

The sum of the firstn odd natural numbers is n2.

Question 6 :

Find the sum of all odd numbers between 100 and 200

Answer 6 :

The series is 101,103, 105, …, 199

Let the number ofterms be n

So, a = 101, d = 103 –101 = 2, an = 199

an = a+ (n-1)d

199 = 101 + (n-1)2

199 = 101 + 2n – 2

2n = 199 – 101 + 2

2n = 100

n = 100/2

= 50

By using the formula,

The sum of n terms = S= n/2[a + l]

= 50/2 [101 + 199]

= 25 [300]

= 7500

The sum of the oddnumbers between 100 and 200 is 7500.

Question 7 : Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667

Answer 7 :

The odd numbersbetween 1 and 1000 divisible by 3 are 3, 9, 15,…,999

Let the number ofterms be ‘n’, so the nth term is 999

a = 3, d = 9-3 = 6, an =999

an = a+ (n-1)d

999 = 3 + (n-1)6

999 = 3 + 6n – 6

6n = 999 + 6 – 3

6n = 1002

n = 1002/6

= 167

By using the formula,

Sum of n terms, S =n/2 [a + l]

= 167/2 [3 + 999]

= 167/2 [1002]

= 167 [501]

= 83667

The sum of all oddintegers between 1 and 1000 which are divisible by 3 is 83667.

Hence proved.

Question 8 : Find the sum of all integers between 84 and 719, which are multiples of 5

Answer 8 :

The series is 85, 90,95, …, 715

Let there be ‘n’ termsin the AP

So, a = 85, d = 90-85= 5, an = 715

an = a+ (n-1)d

715 = 85 + (n-1)5

715 = 85 + 5n – 5

5n = 715 – 85 + 5

5n = 635

n = 635/5

= 127

By using the formula,

Sum of n terms, S =n/2 [a + l]

= 127/2 [85 + 715]

= 127/2 [800]

= 127 [400]

= 50800

The sum of all integersbetween 84 and 719, which are multiples of 5 is 50800.

Question 9 : Find the sum of all integers between 50 and 500 which are divisible by 7

Answer 9 :

The series of integersdivisible by 7 between 50 and 500 are 56, 63, 70, …, 497

Let the number ofterms be ‘n’

So, a = 56, d = 63-56= 7, an = 497

an = a+ (n-1)d

497 = 56 + (n-1)7

497 = 56 + 7n – 7

7n = 497 – 56 + 7

7n = 448

n = 448/7

= 64

By using the formula,

Sum of n terms, S =n/2 [a + l]

= 64/2 [56 + 497]

= 32 [553]

= 17696

The sum of allintegers between 50 and 500 which are divisible by 7 is 17696.

Question 10 : Find the sum of all even integers between 101 and 999

Answer 10 :

We know that all evenintegers will have a common difference of 2.

So, AP is 102, 104,106, …, 998

We know, a = 102, d =104 – 102 = 2, an = 998

By using the formula,

an = a+ (n-1)d

998 = 102 + (n-1)2

998 = 102 + 2n – 2

2n = 998 – 102 + 2

2n = 898

n = 898/2

= 449

By using the formula,

Sum of n terms, S =n/2 [a + l]

= 449/2 [102 + 998]

= 449/2 [1100]

= 449 [550]

= 246950

The sum of all evenintegers between 101 and 999 is 246950.


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RD Chapter 19- Arithmetic Progressions Ex-19.4 Contributors

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