# NCERT Solution Exercise 5.3 Class 10 Maths

Important formula for this Exercise:

Sum of n terms of AP= S

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

(ii) S= n/2[a + an ]

(iii) S= n/2[a + l ]

(iv) Sum of first n positive integers = S= n(n+1)/2

Q.1 Find the sum of the Following APs;

(i) 2, 7,12,…………to 10 terms

a = first term = 2

d = common difference = 7-2 = 5

n = Number if terms = 10

Sum of n terms of AP

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

⇒ S10 = 10/2[2×2 + (10-1)×5]

⇒ S10 = 5[4 + 9×5]

⇒ S10 = 5[4 + 45]

⇒ S10 = 5

⇒ S10 = 245

(ii) -37, -33, -29, …………,to 12 terms

a = First term = 2

d = common difference = -33 – (-37) = -33 + 37 = 4

n = number of terms = 12

Sum of n terms of A.P.

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

⇒ S12 = 12/2[2×-37 + (12-1)×4]

⇒ S12 = 6[2×-37 + 11×4]

⇒ S12 = 6[-74 + 44]

⇒  S12 = 6[-30]

⇒ S12 = -180

(iii) 0.6, 1.7, 2.8, ………….to 100 terms.

a = First terms = 0.6

d = common difference = 1.7 – 0.6 = 1.1

n = number of terms = 100

Sum of n terms in A.P.

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

S100 = 100/2[2×0.6 + (100 -1)×1.1]

⇒ S100 = 50[2×0.6 +99×1.1]

⇒ S100 = 50[1.2 + 108.9]

⇒ S100 = 50×110.1

⇒ S100 = 5505

(iv) 1/15, 1/12, 1/10, ………to 11 terms

a = First terms  = 1/15

d = Common difference = 1/12 – 1/15 = (5-4)/60 = 1/60

n = number of terms = 11

Sum of n terms in AP

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

S11 = 11/2[2×1/15 + (11-1)×1/60]

⇒ S11 = 11/2[2/15 + (10)×1/60]

⇒ S11 = 11/2[2/15 + 1/6]

⇒ S11 = 11/2[2/15 + 1/6]

⇒ S11 = 11/2[(4+5)/30]

⇒ S11 = 11/2[9/30]

⇒ S11 = 11/2[3/10]

⇒ S11 = 33/20

Q.2 Find the Sums given below :

(i) 7+10½+14+ ……………+ 84

Solution:

a = First term = 7

d = Common difference = 10½ – 7 = 21/2 – 7 = (21-14)/2 = 7/2

an = nth term of AP = 84

n =?

Sn =?

an = 84

⇒ a + (n-1)d = 84

⇒ 7+ (n-1)7/2 = 84

⇒ (n-1)7/2 = 84-7

⇒ (n-1) = 77×2/7

⇒ (n-1) = 11×2

⇒ (n-1) = 22

⇒ n = 22+1

⇒ n = 23

Sum of n terms in AP

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

⇒ S23 = 23/2[2×7 + (23-1)×7/2]

⇒ S23 = 23/2[14 + 22×7/2]

⇒ S23 = 23/2[14 + 11×7]

⇒ S23 = 23/2[14 + 77]

⇒ S23 = 23/2

⇒ S23 = 23/2

⇒ S23 = 2093/2

⇒ S23 = 1046½

(ii) 34+32+30+……………+10

Solution:

a = First term = 34

d = 32-34 = -2

an = nth term of AP = 10

⇒ a + (n-1)d = an

⇒ 34 + (n-1)×-2 = 10

⇒ (n-1)×-2 = 10 – 34

⇒ (n-1) = -24/-2

⇒ n-1 = 12

⇒ n = 12+1

⇒ n = 13

Sum of n terms in AP

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

⇒ S13 = 13/2[2×34 + (13-1)×-2]

⇒ S13 = 13/2[68 + 12×-2]

⇒ S13 = 13/2[68 – 24]

⇒ S13 = 13/2

⇒ S13 = 13×44

⇒ S13 = 286

(iii) -5+(-8)+(-11)+……………………+(-230)

Solution:

a = First term =-5

d = Common difference = -8-(-5)= -8+5 = -3

an = nth term of AP = -230

⇒ a + (n-1)d = an

⇒ -5+ (n-1)×-3 = -230

⇒ (n-1)×-3 = -230+5

⇒ n-1 = -225/-3

⇒ n-1 = 75

⇒ n = 75+1

⇒ n = 76

Sum of n terms in AP:

S= n/2[a +an ]

⇒ S76 = 76/2[-5-230]

⇒ S76 = 38[-235]

⇒ S76 = -8930

Q.3 In an AP

(i) Given:

a=5,

d=3,

an=50

Find:

n=?

Sn =?

Solution:

an=50

a+(n-1)d = an

⇒ 5+(n-1)×3 = 50

⇒ (n-1)×3 = 50-5

⇒ n-1 = 45/3

⇒ n-1 = 15

⇒ n = 15+1

⇒ n = 16

S= n/2[a + an ]

⇒ S16 = 16/2[5 + 50 ]

⇒ S16 = 8[55 ]

⇒ S16 = 440

(ii) Given:

a = 7

a13 =35

Find:

d=?

s13 =?

Solution:

a13 =35

⇒ a+12d = a13

⇒ a+12d = 35

⇒ 7+12d = 35

⇒ 12d = 35-7

⇒ 12d = 28

⇒ d = 28/12

⇒ d = 7/3

S= n/2[a +an ]

⇒ S13 = 13/2[7 + 35]

⇒ S13 = 13/2

⇒ S13 = 13×21

⇒ S13 = 273

(iii) Given:

a12 = 37

d = 3

Find:

a =?

a12 =?

Solution:

a+11d = a12

⇒ a + 11d = 37

⇒ a + 11×3 = 37

⇒ a + 33 = 37

⇒ a = 37 – 33

⇒ a = 4

S= n/2[a + an ]

⇒ S12 = 12/2[4 + 37 ]

⇒ S12 = 6

⇒ S12 = 246

(iv) Given:

a3 = 15

s10 = 125

To find:

d=?

a10 =?

Solution:

a3 =15

⇒ a+2d = 15 ———(i)

Sum of n terms:

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

⇒   10/2[2a + (10-1)d] =S10

⇒ 5[2a + 9d] = 125

⇒ [2a + 9d] = 125/5

⇒ 2a + 9d = 25 —————–(ii)

By Elimination method:

a+2d = 15 ———(i)×2

2a + 9d = 25 ——(ii)

[ After multiplying equation no (i) by 2 ]

2a + 4d = 30 ——(iii)

2a + 9d = 25 ——(ii)

Subtracting Equation(ii) From Equation (iii)

-5d = 5

⇒ d = 5/-5

⇒ d = -1 Put into Equation (i)

a+2d = 15

⇒ a + 2×-1 = 15

⇒ a -2 = 15

⇒ a = 15+2

⇒ a = 17

a10 = a+9d

⇒ a10 = a + 9d

⇒ a10 = 17 + 9×-1

⇒ a10 = 17 – 9

⇒ a10 = 8

(v) Given:

d = 5

S= 75

To Find:

a =?

a=?

Solution:

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

⇒  n/2[2a+(n-1)d] = S9

⇒ 9/2[2a +(9-1)5] = 75

⇒ 9/2 [ 2a + 8×5] = 75

⇒ 2a+40 = 75×2/9

⇒ 2(a+20) = 75×2/9

⇒ a+20 = 75×2/2×9

⇒ a+20 = 25/3

⇒ a = 25/3 -20

⇒ a = -35/3

a= a+8d

⇒ a= -35/3 + 8×5

⇒ a= -35/3 + 40

⇒ a= (-35+120)/3

⇒ a= 85/3

(vi) Given:

a = 2

d = 8

sn = 90

To Find:

n = ?

an =?

Solution:

Sn = 90

⇒ n/2[2a+(n-1)d] = Sn

⇒ n/2[2×2+(n-1)8] = 90

⇒ n/2[4+8n-8] = 90

⇒ n/2 × [8n -4] = 90

⇒ n/2 × 2[4n -2] = 90

⇒ n[4n-2] = 90

⇒ 4n² – 2n – 90 = 0

⇒ 2(2n² – n – 45) = 0

⇒ 2n² – n – 45 = 0

⇒ 2n² – (10-9)n – 45 = 0

⇒ 2n² – 10n + 9n – 45 = 0

⇒ 2n(n-5) + 9(n-5) = 0

⇒ (n-5)(2n + 9) = 0

n-5 = 0 ; 2n + 9 = 0

⇒ n = 5; 2n = -9

⇒ n = 5; n = -9/2

an = a+(n-1)d

⇒ a5 = 2+(5-1)8

⇒ a5 = 2+4×8

⇒ a5 = 2+32

⇒ a5 = 34

(vii) Given:

a = 8

an = 62

S=210

To Find:

n =?

an =?

Solution:

n/2[a + an]= Sn

n/2[8+62] = 210

⇒ n/2 [ 70] = 210

⇒ n×35 = 210

⇒ n = 210/35

⇒ n = 6

an = 62

8 + (6-1)d = 62

⇒ 8+5d = 62-8

⇒  5d= 54

⇒ d= 54/5

(viii) Given:

an = 4

d = 2

S= -14

To Find:

n =?

a =?

Solution:

an = 4

⇒ a+(n-1)d = an

⇒ a + (n-1)2 = 4

⇒ a + 2n -2 = 4

⇒ a + 2n = 4 + 2

⇒ a + 2n = 6

⇒ a = 6 – 2n ——(i)

n/2[a + an] = S

⇒ n/2[a+4] = -14

⇒ n/2[6-2n+4 ] = -14  [ from equation (i) ]

⇒ n/2[10-2n] = -14

⇒ n[5-n] = -14

⇒ 5n – n² = -14

⇒ n²- 5n – 14 = 0

⇒ n² – (7-2)n -14 = 0

⇒ n² – 7n + 2n -14 = 0

⇒ n[n – 7] + 2[n-7] = 0

⇒ [n-7][n+2] = 0

n-7 = 0 ; n+2 = 0

⇒ n = 7 ; n = -2(not possible)

⇒ n = 7 Put into equation (i)

a = 6 – 2n ——(i)

⇒ a = 6 – 2×7

⇒ a = 6 – 14

⇒ a = -8

(ix) Given:

a = 3

n = 8

s = 192

To Find:

d = ?

Solution:

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

⇒ 8/2[2×3 + (8-1)d ] = 192

⇒ 4 [6+7d ] = 192

⇒ 6 + 7d = 192/4

⇒ 7d = 48

⇒ 7d = 48-6

⇒ 7d = 42

⇒ d = 42/7

⇒ d = 6

(x) Given:

l = 28

s= 144

n = 9

To find:

a=?

Solution:

n/2[a +l] = Sn

9/2[ a+28] = 144

⇒ a+18 =144×2/9

⇒ a +28 = 16×2

⇒  a =32 -28

⇒ a = 4

Q4. How many terms of the AP: 9,17,25,……..must be taken to given a sim of 636?

Solution:

AP: 9,17,25,…..

a = 1st term = 9

d = common difference = 17-9 = 8

Sn = Sum of n terms =636

n = Number of terms =?

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

⇒ n/2 [2×9 + (n-1)×8] = 636

⇒ n/2 [18 + (n-1)×8] = 636

⇒ n/2 [18 + 8n -8 ] = 636

⇒ n/2 [10 + 8n ] = 636

⇒ n[5 + 4n] = 636

⇒ 5n + 4n² = 636

⇒ 4n² + 5n – 636 =0

⇒ 4n² +(53-48)n – 636 = 0

⇒ 4n² +53n – 48n – 636 = 0

⇒ n(4n + 53) – 12(4n + 53) = 0

⇒ (4n + 53)(n-12) = 0

4n + 53 = 0; n -12 = 0

⇒ 4n = -53 ; n = 12

⇒ n = -53/4(not possible) ;

Hence Number of terms 12

Q5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Solution:

a = 1st term =5

l = last term  = 45

Sn = Sum of n terms =400

n = Number of terms =?

d = Common difference = ?

Sum of n terms of AP:

n/2[a + l] = Sn

⇒ n/2[5+45] = 400

⇒ n/2 [ 50 ] = 400

⇒ n×25 = 400

⇒ n = 400/25

⇒ n = 16

last term = 45

16th term  = 45

a+(n-1)d = Last term(16th term)

⇒ 5+[16-1]d = 45

⇒ 5+ d = 45

⇒ 15d = 45 -5

⇒ 15d = 40

⇒ d = 40/15

⇒ d = 8/3

Q6. The first and the last terms of an AP are 17 and 35 respectively. If the common difference is 9, how many terms are there and what us their sum ?

Solution:

a = 1st term =5

an= l = last term  = 45

Sn = Sum of n terms =?

a + (n-1)d = an

⇒ 17 + (n-1)9 = 350

⇒ (n-1)9 = 350 – 17

⇒ (n-1) = 333/9

⇒ n -1 = 37

⇒ n = 37 +1

⇒ n = 38

Sn = n/2[a + l]

S38 = 38/2[17+350]

⇒ S38 = 19

⇒ S38 = 6973

Q7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

Solution:

d = 7

a22 = 149

s22 = ?

a22 = 149

⇒ a + 21d = 149

⇒ a + 21×7 = 149

⇒ a + 147 = 149

⇒ a = 149 – 147

⇒ a = 2

S= n/2[a + an]

⇒ S22  = 22/2[a + a22]

⇒  S22  = 11[2 + 149]

⇒ S22  = 11

⇒ S22  = 1661

Hence sum of 1st 22 term of AP = 1661

Q8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Solution:

2nd term = 14

⇒ a= 14

⇒ a + d = 14 ——-(i)

3rd term = 18

⇒ a3 = 18

⇒ a + 2d = 18 ——–(ii)

By Elimination method

a + d = 14 ——-(i)

a + 2d = 18 ——–(ii)

–    –      = –                [ By subtracting equation (i) From equation (ii) ]

————————–

⇒ – d = – 4

⇒ d = 4 put into equation (i)

a + d = 14

⇒ a + 4 = 14

⇒ a = 14 – 4

⇒ a = 10

here a = 10 , d = 4,

Sn = S51 =?

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

S51 = 51/2[2×10+(51-1)×4]

⇒ S51 = 51/2[20+(50)×4]

⇒ S51 = 51/2[ 20+200 ]

⇒ S51 = 51/2

⇒ S51 = 51×110

⇒ S51 = 5610

Hence sum of 1st 51terms = 5610

Q.9 If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, Find the sum of first n terms.

Solution:

Sum of first 7 terms of AP = 49

⇒ S7 = 49

17th term = 289

Sum of 1st n terms = Sn =?

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

⇒ 7/2[2a+(7-1)d = S7

⇒ 7/2[2a+6d] = 49

⇒ 7[a+3d] = 49

⇒ a+3d = 49/7

⇒ a+3d = 7 ———(i)

17th term = 289

⇒ S17 = 289

⇒ 17/2[2a+16d] = 289

⇒ 17[a+8d] = 289

⇒ a+8d = 289/17

⇒ a+8d = 17 ———(ii)

By Elimination Method:

a + 3d = 7 ———–(i)

a + 8d = 17 ———-(ii)

–    –      = –              [ subtracting (i) from equation(ii) ]

-5d = – 10

⇒ d = -10/-5

⇒ d = 2 Put into equation(i)

a + 3d = 7 ———–(i)

a + 3×2 = 7

⇒ a + 6 = 7

⇒ a = 7-6

⇒ a = 1

Q 10. Show that a1,a2,…………..an ………………….form an AP where a is defined as below:

(i) an  = 3 + 4n                                             (ii) an = 9 – 5n

Also find the sum of the first 15 terms in each case.

Solution:

(i) an  = 3 + 4n

Put n =1

a1 = 3 + 4×1 = 7

Put n = 2

a2 = 3 + 4×2 = 11

Put n =3

a3 = 3 + 4×3 = 15

AP: 7, 11, 15, ………….

a = First term = 7

d = Common difference = 11-7 =  4

Sum of First 15 terms

S15 = 15/2[2a+(15-1)d]

⇒ S15 = 15/2[2×7 + (14)×4]

⇒ S15 = 15/2[14 + 56]

⇒ S15 = 15/2

⇒ S15 = 15×35

⇒ S15 = 525

(ii) an = 9 – 5n

Put n=1

a1 = 9 – 5×1 = 4

Put n = 2

a2 = 9 – 5×2 = -1

Put n = 3

a2 = 9 – 5×3 = -6

Put n = 4

a4 = 9 – 5×4 = -11

AP: 4, -1, -6, ………………

a = First term = 4

d = Common difference = -1-4 = -5

Sum of 15th terms

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

⇒ S15 = 15/2[2×4+(15-1)×-5]

⇒ S15 = 15/2[2×4+(14)×-5]

⇒ S15 = 15/2[8 – 70]

⇒ S15 = 15/2[-62]

⇒ S15 = 15×-31

⇒ S15 = -465

Q.11 If the sum of the first n terms of an AP is 4n – n², What is first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, Find the third the 10th and the  nth terms.

Given:

Sum of First n terms of an AP S= 4n – n²

To Find:

(i) First term (S1) =?

(ii) Sum of 1st two terms S2 = ?

(iii) 2nd term a2 =?

(iv) 3rd term a=?

(v) 10th term a=?

(vi) nth term an =?

Solution:

(i) S= 4n – n²

Put n = 1

S1 = 4×1 – 1²

⇒ S1 = 4 – 1 = 3

(ii)Put n = 2

⇒ S2 = 4×2 – 2²

⇒ S2 = 8 – 4

⇒ S2 = 4

Put n = 3

S3 = 4×3 – 3²

⇒ S3 = 12 – 9 = 3

(iii) a= Sn– Sn-1

⇒ a2 = S2– S2-1

⇒ a2 = S2 – S1

⇒ a2 = 4 – 3

⇒ a2 = 1

(iv) a3 = S3 – S3-1

⇒ a3 = S3 – S2

⇒ a3 = 3 – 4

⇒ a3 = -1

AP: 3, 1, -1, ………

a = 3

d = 1-3 = -2

a10 = a+9d

⇒ a10 = 3+9×-2

⇒ a10 = 3 – 18

⇒ a10 = -15

an = a+(n-1)d

⇒ an = 3+(n-1)×-2

⇒ an = 3 – 2n +2

⇒ an = 5 – 2n

Hence:

(i) First term (S1) = 3

(ii) Sum of 1st two terms S2 = 4

(iii) 2nd term a2 = 1

(iv) 3rd term a=-1

(v) 10th term a10 = -15

(vi) nth term an = 5 – 2n

Q. 12. Find the sum of the first 40 positive integers divisible by 6.

Solution:

AP: 6, 12, 18, ……………to 40 terms

a = 6

d = 12-6 = 6

n = 40

Sum of n terms

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

⇒ S40 = 40/2[2×6 + (40-1)×6]

⇒ S40 = 20[12 + 39×6]

⇒ S40 = 20[12 + 234]

⇒ S40 = 20

⇒ S40 = 4920

Hence sum of 1st 40 positive integers = 4920

Q.13. Find the sum of the first 15 multiples of 8.

Solution:

AP: 8, 16, 24, ……………..to 15 terms

a = 8

d = 16 – 8 = 8

Sum of n terms of AP

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

⇒ S15 = 15/2[2×8+(15-1)×8]

⇒ S15 = 15/2[16+(14)×8]

⇒ S15 = 15/2[16+112]

⇒ S15 = 15/2

⇒ S15 = 15×64

⇒ S15 = 960

Hence sum of first 15 multiples of 8 = 960

Q.14. Find the sum of the odd numbers between 0 and 50.

Solution:

0,1,2,3,4,……………………………………..48,49,50

AP: 1, 3, 4, ……………………… 49

a = 1st term = 1

d = Common difference = 3-1 = 2

an = nth term of AP = 49

n =?

Sn = Sum of n terms =?

an = 49

⇒ a+(n-1)d = 49

⇒ 1+(n-1)×2 = 49

⇒ (n-1)×2 = 49-1

⇒ (n-1)×2 = 48

⇒ n-1 = 48/2

⇒ n-1 = 24

⇒ n = 24+1

⇒ n = 25

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

⇒ S25 = 25/2[2×1+(25-1)×2]

⇒ S25 = 25/2[2+(24)×2]

⇒ S25 = 25/2[2+48]

⇒ S25 = 25/2

⇒ S25 = 25×25

⇒ S25 = 625

Q.15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows:₹200 for the first day, ₹ 250 for the second day, ₹300 for the third day etc. the penalty for each succeeding day being ₹50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days.

Solution:

AP: 200, 250, 300, …………….to 30 terms.

a = 1st term = 200

d = common difference = 250 – 200 = 50

n = number of terms = 30

Sn = Sum of n terms = ?

Sn = n/2[2a+(n – 1)d]

⇒ S30 = 30/2[2×200 + (30-1)×50]

⇒ S30 = 30/2[400 + 29×50]

⇒ S30 = 15[400 + 1450]

⇒ S30 = 15

⇒ S30 = 27750

Hence total money the contractor has to pay as penalty = ₹ 27750

Q.16. A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prizes is ₹ 20 less than its preceding prize. Find the value of each of the prizes.

Given:

Sn = 700

n = 7

d = -20

To find:

First Prize = a1 =?

Second Prize =a2 =?

Third Prize = a3 =?

Fourth Prize = a4 =?

Fifth Prize = a5 =?

6th Prize = a6 =?

7th Prize = a7 =?

Solution:

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

⇒ 7/2[2a+(7-1)×-20] = S7

⇒ 7/2[2a+(6)×-20] = 700

⇒ 7/2[2a-120] = 700

⇒ 7[a -60] = 700

⇒ a – 60 = 700/7

⇒ a – 60 = 100

⇒ a = 100 + 60

⇒ a = 160

a2 = a + d =160-20 = 140

a3 = a2+d = 140 – 20 =120

a4  = a+ d = 120 – 20 = 100

a= a4 + d = 100 – 20 = 80

a= a5 + d = 80 – 20 = 60

a= a6 + d = 60 – 20 = 40

Hence Each cash prizes are:

First Prize = a1 = 160

Second Prize =a2 = 140

Third Prize = a3 = 120

Fourth Prize = a4 = 100

Fifth Prize = a5 = 80

6th Prize = a6 = 60

7th Prize = a7 = 40

Q.17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class 1st will Plant 1 tree, a section of Class II will Plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

Solution:

AP: 3, 6, 9, ……………………………………..36

a = 1st term = 3

d = Common difference = d = 6-4 = 3

an = Last class(12th) will plant = 36

n = 12

Sn = Total number of tree will plant = ?

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

⇒ S12 = 12/2[2×3 + (12-1)×3]

⇒ S12 = 6[6 + 11×3]

⇒ S12 = 6[6 + 33]

⇒ S12 = 6

⇒ S12 = 234

Hence total number of trees will planted = 234

Q.18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, …… as shown in Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive semicircles?(Take π = 22/7) Solution:

Width of

l1 = Length of  First spiral = πr =π0.5 = 5π/10 = π/2

l2 = Length of  First spiral = πr =π×1.0 = π

l3 = Length of  First spiral = πr =π×1.5 = 15π/10 = 3π/2

l4 = Length of  First spiral = πr =π×2.0 = 2π

AP: π/2, π,  2π , ……… , 13π/2

a = π/2

d =π –  π/2 = π/2

n = 13( total number of spiral) = ?

Sn = n/2[ a + an ]

⇒ S13 = 13/2[ π/2 + 13π/2 ]

⇒ S13 = 13/2[14π/2]

⇒ S13 = 13/2 × 14π/2

⇒ S13 = 13/2 × 7×22/7

⇒ S13 = 143

Q.19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row 18 in the row next to it and so on (see fig. 5.5). In how many rows are the 200 logs placed and how many logs are in the top row? Given:

Total number of logs = 200 = Sn

AP: 20, 19, 18, ……..

To Find:

(i) Total number of Rows = n = ?

(ii) How many logs in top row = an = ?

Solution:

a = First term = 20

d = Common difference = -1

n/2[ 2a +(n-1) d ] =Sn

⇒ n/2[ 2×20 + (n-1)×-1 ] = 200

⇒ n/2 [40 – n + 1)] = 200

⇒ n/2[41 – n ] = 200

⇒ n[41 – n ] = 200×2

⇒ 41n – n² = 400

⇒ n² – 41n + 400 = 0

⇒ n² – (25 + 16)n + 400 = 0

⇒ n² – 25n – 16n + 400 = 0

⇒ n(n – 25) – 16(n – 25) = 0

⇒ (n-25)(n-16) = 0

n – 25 = 0 ; n -16 = 0

⇒ n = 25 ; n = 16

If n = 25

The Number of logs in top (25th) row = a25 = a + 24d

⇒ a25 = 20 + 24×-1

⇒  a25 = 20 – 24

⇒  a25 = -4 which is not possible

If n = 16

The Number of logs in top (16th) row = a16 = a + 15d

⇒ a16 = 20 + 15×-1

Hence

(i) total number of rows = 16

(ii) Number of logs in top(16th) row = 5

Q.20 In a potato race , a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line(see Fig.5.6) A competitor starts from the bucket, picks up the nearest potato, runs back with it , drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

Solution:

AP : 10, 16, 22, ……………….

a = First term = 10

d = Common difference = 16 – 10 = 6

n = Total number of potatoes = 10

Total distance  Sn =?

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

⇒ S10 = 10/2[2×10+(10-1)×6]

⇒ S10 = 5[2×10+9×6]

⇒ S10 = 5[20 +54]

⇒ S10 = 5

⇒ S10 = 370

Hence Total distance run by the competitor = 370m